TG 10th Quadratic Equations

10th Maths Quadratic Equations Exercise 5.4 Solutions

zaheerpashamd

Exercise 5.4 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on the nature of roots of quadratic equations and applications to word problems. Mathematical expressions are rendered using MathJax.

1. Find the nature of the roots of the following quadratic equations. If real roots exist, find them.

(i) \( 2x^2 – 3x + 5 = 0 \)

For \( ax^2 + bx + c = 0 \), nature of roots depends on the discriminant: \( \Delta = b^2 – 4ac \).

Here, \( a = 2 \), \( b = -3 \), \( c = 5 \).

Discriminant: \( \Delta = (-3)^2 – 4(2)(5) = 9 – 40 = -31 \).

Since \( \Delta < 0 \), the roots are not real (they are complex).

Nature of roots: Not real (complex)

(ii) \( 3x^2 – 4\sqrt{3}x + 4 = 0 \)

\( a = 3 \), \( b = -4\sqrt{3} \), \( c = 4 \).

Discriminant: \( \Delta = (-4\sqrt{3})^2 – 4(3)(4) = 16 \cdot 3 – 48 = 48 – 48 = 0 \).

Since \( \Delta = 0 \), the roots are real and equal.

Root: \( x = \frac{-b}{2a} = \frac{4\sqrt{3}}{2 \cdot 3} = \frac{2\sqrt{3}}{3} \).

Check: \( 3\left(\frac{2\sqrt{3}}{3}\right)^2 – 4\sqrt{3}\left(\frac{2\sqrt{3}}{3}\right) + 4 = 3 \cdot \frac{12}{9} – 4\sqrt{3} \cdot \frac{2\sqrt{3}}{3} + 4 = 4 – \frac{8 \cdot 3}{3} + 4 = 4 – 8 + 4 = 0 \).

Nature of roots: Real and equal, Root: \( x = \frac{2\sqrt{3}}{3} \)

(iii) \( 2x^2 – 6x + 3 = 0 \)

\( a = 2 \), \( b = -6 \), \( c = 3 \).

Discriminant: \( \Delta = (-6)^2 – 4(2)(3) = 36 – 24 = 12 \).

Since \( \Delta > 0 \), the roots are real and distinct.

Roots: \( x = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2} \).

Check: \( x = \frac{3 + \sqrt{3}}{2} \), \( 2\left(\frac{3 + \sqrt{3}}{2}\right)^2 – 6\left(\frac{3 + \sqrt{3}}{2}\right) + 3 = 2 \cdot \frac{9 + 6\sqrt{3} + 3}{4} – 3(3 + \sqrt{3}) + 3 = \frac{12 + 6\sqrt{3}}{2} – 9 – 3\sqrt{3} + 3 = 0 \).

Nature of roots: Real and distinct, Roots: \( x = \frac{3 \pm \sqrt{3}}{2} \)

2. Find the values of \( k \) for each of the following quadratic equations, so that they have two equal roots.

(i) \( 2x^2 + kx + 3 = 0 \)

For equal roots, \( \Delta = 0 \). Here, \( a = 2 \), \( b = k \), \( c = 3 \).

Discriminant: \( \Delta = k^2 – 4(2)(3) = k^2 – 24 \).

Set \( \Delta = 0 \): \( k^2 – 24 = 0 \implies k^2 = 24 \implies k = \pm \sqrt{24} = \pm 2\sqrt{6} \).

Value of \( k \): \( k = \pm 2\sqrt{6} \)

(ii) \( kx(x – 2) + 6 = 0 \), \( k \neq 0 \)

Expand: \( kx^2 – 2kx + 6 = 0 \).

\( a = k \), \( b = -2k \), \( c = 6 \).

Discriminant: \( \Delta = (-2k)^2 – 4(k)(6) = 4k^2 – 24k \).

Set \( \Delta = 0 \): \( 4k^2 – 24k = 0 \implies 4k(k – 6) = 0 \implies k = 0 \) or \( k = 6 \). Since \( k \neq 0 \), \( k = 6 \).

Value of \( k \): \( k = 6 \)

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

Let the breadth be \( x \) m, length = \( 2x \) m.

Area: \( x \cdot 2x = 800 \implies 2x^2 = 800 \implies x^2 = 400 \).

Quadratic: \( x^2 – 400 = 0 \).

Discriminant: \( \Delta = 0 – 4(1)(-400) = 1600 \).

Since \( \Delta > 0 \), real roots exist, so it is possible.

Roots: \( x = \pm \sqrt{400} = \pm 20 \). Take \( x = 20 \).

Breadth: \( 20 \) m, length: \( 40 \) m.

Check: Area = \( 40 \cdot 20 = 800 \) m².

Possible: Yes, Length: 40 m, Breadth: 20 m

4. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the above situation possible? If so, determine their present ages.

Let their present ages be \( x \) and \( y \).

Sum: \( x + y = 20 \implies y = 20 – x \).

Four years ago: Ages = \( x – 4 \), \( y – 4 \).

Product: \( (x – 4)(y – 4) = 48 \).

Substitute \( y \): \( (x – 4)(20 – x – 4) = 48 \implies (x – 4)(16 – x) = 48 \).

Expand: \( 16x – x^2 + 64 – 16x = 48 \implies -x^2 + 64 = 48 \implies x^2 – 16 = 0 \).

Discriminant: \( \Delta = 0 – 4(1)(-16) = 64 \).

Since \( \Delta > 0 \), real roots exist.

Roots: \( x = \pm \sqrt{16} = \pm 4 \). Take \( x = 4 \), \( y = 16 \).

Ages 4 years ago: \( 0 \) and \( 12 \), product = \( 0 \cdot 12 = 0 \), not 48. The discriminant suggests real roots, but ages must be positive and yield a product of 48 four years ago, which isn’t satisfied here.

Possible: No

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth. Comment on your answer.

Let length be \( x \) m, breadth be \( y \) m.

Perimeter: \( 2(x + y) = 80 \implies x + y = 40 \implies y = 40 – x \).

Area: \( x y = 400 \).

Substitute: \( x (40 – x) = 400 \implies 40x – x^2 = 400 \implies x^2 – 40x + 400 = 0 \).

Discriminant: \( \Delta = (-40)^2 – 4(1)(400) = 1600 – 1600 = 0 \).

Since \( \Delta = 0 \), real and equal roots exist, so it is possible.

Root: \( x = \frac{40}{2} = 20 \), \( y = 20 \).

Check: Perimeter = \( 2(20 + 20) = 80 \), Area = \( 20 \cdot 20 = 400 \).

Comment: The park is a square (length = breadth), which is a special case of a rectangle.

Possible: Yes, Length: 20 m, Breadth: 20 m, Comment: The park is a square

10th Maths Quadratic Equations Exercise 5.3 Solutions

zaheerpashamd

Exercise 5.3 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on solving quadratic equations using the quadratic formula and applying them to word problems. Mathematical expressions are rendered using MathJax.

1. Find the roots of the following quadratic equations, if they exist.

(i) \( 2x^2 + x – 4 = 0 \)

Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), \( c = -4 \).

Discriminant: \( \Delta = b^2 – 4ac = 1^2 – 4(2)(-4) = 1 + 32 = 33 \).

Since \( \Delta > 0 \), real roots exist.

Roots: \( x = \frac{-1 \pm \sqrt{33}}{4} \).

Check: \( x = \frac{-1 + \sqrt{33}}{4} \), \( 2\left(\frac{-1 + \sqrt{33}}{4}\right)^2 + \frac{-1 + \sqrt{33}}{4} – 4 = \frac{2(1 – 2\sqrt{33} + 33)}{16} + \frac{-1 + \sqrt{33}}{4} – 4 = \frac{34 – 2\sqrt{33}}{8} + \frac{-1 + \sqrt{33}}{4} – 4 \). This simplifies to 0 (numerically verified).

Roots: \( x = \frac{-1 \pm \sqrt{33}}{4} \)

(ii) \( 4x^2 + 4\sqrt{3}x + 3 = 0 \)

\( a = 4 \), \( b = 4\sqrt{3} \), \( c = 3 \).

Discriminant: \( \Delta = (4\sqrt{3})^2 – 4(4)(3) = 48 – 48 = 0 \).

Since \( \Delta = 0 \), one real root exists.

Root: \( x = \frac{-b}{2a} = \frac{-4\sqrt{3}}{2 \cdot 4} = -\frac{\sqrt{3}}{2} \).

Check: \( 4\left(-\frac{\sqrt{3}}{2}\right)^2 + 4\sqrt{3}\left(-\frac{\sqrt{3}}{2}\right) + 3 = 4 \cdot \frac{3}{4} – 4 \cdot \frac{3}{2} + 3 = 3 – 6 + 3 = 0 \).

Root: \( x = -\frac{\sqrt{3}}{2} \) (repeated)

(iii) \( 5x^2 – 7x – 6 = 0 \)

\( a = 5 \), \( b = -7 \), \( c = -6 \).

Discriminant: \( \Delta = (-7)^2 – 4(5)(-6) = 49 + 120 = 169 \).

Since \( \Delta > 0 \), real roots exist.

Roots: \( x = \frac{7 \pm \sqrt{169}}{2 \cdot 5} = \frac{7 \pm 13}{10} \).

\( x = \frac{20}{10} = 2 \), \( x = \frac{-6}{10} = -\frac{3}{5} \).

Check: \( x = 2 \): \( 5(2)^2 – 7(2) – 6 = 20 – 14 – 6 = 0 \), \( x = -\frac{3}{5} \): \( 5\left(-\frac{3}{5}\right)^2 – 7\left(-\frac{3}{5}\right) – 6 = 5 \cdot \frac{9}{25} + \frac{21}{5} – 6 = \frac{9}{5} + \frac{21}{5} – 6 = 6 – 6 = 0 \).

Roots: \( x = 2, -\frac{3}{5} \)

(iv) \( x^2 + 5 = -6x \)

Rewrite: \( x^2 + 6x + 5 = 0 \).

\( a = 1 \), \( b = 6 \), \( c = 5 \).

Discriminant: \( \Delta = 6^2 – 4(1)(5) = 36 – 20 = 16 \).

Since \( \Delta > 0 \), real roots exist.

Roots: \( x = \frac{-6 \pm \sqrt{16}}{2} = \frac{-6 \pm 4}{2} \).

\( x = \frac{-2}{2} = -1 \), \( x = \frac{-10}{2} = -5 \).

Check: \( x = -1 \): \( (-1)^2 + 5 + 6(-1) = 1 + 5 – 6 = 0 \), \( x = -5 \): \( (-5)^2 + 5 + 6(-5) = 25 + 5 – 30 = 0 \).

Roots: \( x = -1, -5 \)

2. Find the roots of the quadratic equations given in Q.1 by applying the quadratic formula.

The roots for Q.1 have already been found using the quadratic formula. Refer to the solutions above:

(i) \( x = \frac{-1 \pm \sqrt{33}}{4} \), (ii) \( x = -\frac{\sqrt{3}}{2} \), (iii) \( x = 2, -\frac{3}{5} \), (iv) \( x = -1, -5 \).

3. Find the roots of the following equations:

(i) \( x – \frac{1}{x} = 3 \), \( x \neq 0 \)

Multiply by \( x \): \( x^2 – 1 = 3x \implies x^2 – 3x – 1 = 0 \).

\( a = 1 \), \( b = -3 \), \( c = -1 \).

Discriminant: \( \Delta = (-3)^2 – 4(1)(-1) = 9 + 4 = 13 \).

Roots: \( x = \frac{3 \pm \sqrt{13}}{2} \).

Check: \( x = \frac{3 + \sqrt{13}}{2} \), \( \frac{3 + \sqrt{13}}{2} – \frac{2}{3 + \sqrt{13}} = \frac{(3 + \sqrt{13})^2 – 4}{2(3 + \sqrt{13})} = \frac{9 + 6\sqrt{13} + 13 – 4}{2(3 + \sqrt{13})} = 3 \).

Roots: \( x = \frac{3 \pm \sqrt{13}}{2} \)

(ii) \( \frac{1}{x + 4} – \frac{1}{x – 7} = \frac{11}{30} \), \( x \neq -4, 7 \)

Combine: \( \frac{(x – 7) – (x + 4)}{(x + 4)(x – 7)} = \frac{11}{30} \implies \frac{-11}{(x + 4)(x – 7)} = \frac{11}{30} \).

Simplify: \( -11 \cdot 30 = 11 (x + 4)(x – 7) \implies -30 = x^2 – 3x – 28 \implies x^2 – 3x + 2 = 0 \).

\( a = 1 \), \( b = -3 \), \( c = 2 \).

Discriminant: \( \Delta = (-3)^2 – 4(1)(2) = 9 – 8 = 1 \).

Roots: \( x = \frac{3 \pm 1}{2} \implies x = 2, 1 \).

Check: \( x = 2 \): \( \frac{1}{6} – \frac{1}{-5} = \frac{1}{6} + \frac{1}{5} = \frac{11}{30} \), \( x = 1 \): \( \frac{1}{5} – \frac{1}{-6} = \frac{1}{5} + \frac{1}{6} = \frac{11}{30} \).

Roots: \( x = 1, 2 \)

4. The sum of the reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is \( \frac{1}{3} \). Find his present age.

Let Rehman’s present age be \( x \) years.

3 years ago: \( x – 3 \), 5 years from now: \( x + 5 \).

Sum of reciprocals: \( \frac{1}{x – 3} + \frac{1}{x + 5} = \frac{1}{3} \).

Simplify: \( \frac{(x + 5) + (x – 3)}{(x – 3)(x + 5)} = \frac{1}{3} \implies \frac{2x + 2}{x^2 + 2x – 15} = \frac{1}{3} \).

Cross-multiply: \( 3(2x + 2) = x^2 + 2x – 15 \implies 6x + 6 = x^2 + 2x – 15 \implies x^2 – 4x – 21 = 0 \).

\( a = 1 \), \( b = -4 \), \( c = -21 \).

Discriminant: \( \Delta = (-4)^2 – 4(1)(-21) = 16 + 84 = 100 \).

Roots: \( x = \frac{4 \pm \sqrt{100}}{2} = \frac{4 \pm 10}{2} \implies x = 7, -3 \). Take \( x = 7 \).

Check: \( x = 7 \), ages: \( 4 \) and \( 12 \), \( \frac{1}{4} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3} \).

Present age: 7 years

5. In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.

Let marks in Mathematics be \( x \), English be \( y \).

\( x + y = 30 \implies y = 30 – x \).

New marks: Mathematics = \( x + 2 \), English = \( y – 3 \).

Product: \( (x + 2)(y – 3) = 210 \).

Substitute \( y \): \( (x + 2)(30 – x – 3) = 210 \implies (x + 2)(27 – x) = 210 \).

Expand: \( 27x – x^2 + 54 – 2x = 210 \implies -x^2 + 25x + 54 – 210 = 0 \implies x^2 – 25x + 156 = 0 \).

\( a = 1 \), \( b = -25 \), \( c = 156 \).

Discriminant: \( \Delta = (-25)^2 – 4(1)(156) = 625 – 624 = 1 \).

Roots: \( x = \frac{25 \pm 1}{2} \implies x = 13, 12 \).

If \( x = 13 \), \( y = 17 \). If \( x = 12 \), \( y = 18 \).

Check: \( x = 13 \), \( (13 + 2)(17 – 3) = 15 \cdot 14 = 210 \).

Marks: Mathematics = 13, English = 17 (or vice versa)

6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Let the shorter side be \( x \) m, longer side = \( x + 30 \), diagonal = \( x + 60 \).

By Pythagoras: \( x^2 + (x + 30)^2 = (x + 60)^2 \).

Expand: \( x^2 + x^2 + 60x + 900 = x^2 + 120x + 3600 \).

Simplify: \( 2x^2 + 60x + 900 – x^2 – 120x – 3600 = 0 \implies x^2 – 60x – 2700 = 0 \).

\( a = 1 \), \( b = -60 \), \( c = -2700 \).

Discriminant: \( \Delta = (-60)^2 – 4(1)(-2700) = 3600 + 10800 = 14400 \).

Roots: \( x = \frac{60 \pm \sqrt{14400}}{2} = \frac{60 \pm 120}{2} \implies x = 90, -30 \). Take \( x = 90 \).

Shorter side: \( 90 \) m, longer side: \( 120 \) m, diagonal: \( 150 \) m.

Check: \( 90^2 + 120^2 = 8100 + 14400 = 22500 = 150^2 \).

Sides: Shorter = 90 m, Longer = 120 m

7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Let the larger number be \( x \), smaller number be \( y \).

\( x^2 – y^2 = 180 \), \( y^2 = 8x \).

Substitute: \( x^2 – 8x = 180 \implies x^2 – 8x – 180 = 0 \).

\( a = 1 \), \( b = -8 \), \( c = -180 \).

Discriminant: \( \Delta = (-8)^2 – 4(1)(-180) = 64 + 720 = 784 \).

Roots: \( x = \frac{8 \pm \sqrt{784}}{2} = \frac{8 \pm 28}{2} \implies x = 18, -10 \).

Take \( x = 18 \): \( y^2 = 8 \cdot 18 = 144 \implies y = \pm 12 \).

Check: \( 18^2 – 12^2 = 324 – 144 = 180 \), \( 12^2 = 8 \cdot 18 \).

Numbers: 18 and 12 (or \( 18 \) and \( -12 \))

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Let the speed be \( x \) km/h.

Time at speed \( x \): \( \frac{360}{x} \), at \( x + 5 \): \( \frac{360}{x + 5} \).

\( \frac{360}{x} – \frac{360}{x + 5} = 1 \implies 360 \left( \frac{x + 5 – x}{x(x + 5)} \right) = 1 \implies \frac{360 \cdot 5}{x(x + 5)} = 1 \implies x^2 + 5x – 1800 = 0 \).

\( a = 1 \), \( b = 5 \), \( c = -1800 \).

Discriminant: \( \Delta = 5^2 – 4(1)(-1800) = 25 + 7200 = 7225 \).

Roots: \( x = \frac{-5 \pm \sqrt{7225}}{2} = \frac{-5 \pm 85}{2} \implies x = 40, -45 \). Take \( x = 40 \).

Check: \( \frac{360}{40} – \frac{360}{45} = 9 – 8 = 1 \).

Speed: 40 km/h

9. Two water taps together can fill a tank in \( 9 \frac{3}{8} \) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

\( 9 \frac{3}{8} = \frac{75}{8} \) hours.

Let the smaller tap take \( x \) hours, larger tap = \( x – 10 \).

Rates: Smaller = \( \frac{1}{x} \), larger = \( \frac{1}{x – 10} \).

Together: \( \frac{1}{x} + \frac{1}{x – 10} = \frac{1}{\frac{75}{8}} \implies \frac{x – 10 + x}{x(x – 10)} = \frac{8}{75} \).

Simplify: \( \frac{2x – 10}{x^2 – 10x} = \frac{8}{75} \implies 75(2x – 10) = 8(x^2 – 10x) \implies 150x – 750 = 8x^2 – 80x \implies 8x^2 – 230x + 750 = 0 \implies 4x^2 – 115x + 375 = 0 \).

\( a = 4 \), \( b = -115 \), \( c = 375 \).

Discriminant: \( \Delta = (-115)^2 – 4(4)(375) = 13225 – 6000 = 7225 \).

Roots: \( x = \frac{115 \pm \sqrt{7225}}{8} = \frac{115 \pm 85}{8} \implies x = 25, \frac{15}{2} \).

If \( x = 25 \), larger tap: \( 15 \). If \( x = \frac{15}{2} \), larger tap: \( -\frac{5}{2} \) (not possible).

Check: \( \frac{1}{25} + \frac{1}{15} = \frac{3 + 5}{75} = \frac{8}{75} \), matches \( \frac{75}{8} \) hours.

Smaller tap: 25 hours, Larger tap: 15 hours

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Let the speed of the passenger train be \( x \) km/h, express train = \( x + 11 \).

Time: Passenger = \( \frac{132}{x} \), express = \( \frac{132}{x + 11} \).

\( \frac{132}{x} – \frac{132}{x + 11} = 1 \implies 132 \left( \frac{x + 11 – x}{x(x + 11)} \right) = 1 \implies \frac{132 \cdot 11}{x(x + 11)} = 1 \implies x^2 + 11x – 1452 = 0 \).

\( a = 1 \), \( b = 11 \), \( c = -1452 \).

Discriminant: \( \Delta = 11^2 – 4(1)(-1452) = 121 + 5808 = 5929 \).

Roots: \( x = \frac{-11 \pm \sqrt{5929}}{2} = \frac{-11 \pm 77}{2} \implies x = 33, -44 \). Take \( x = 33 \).

Passenger: \( 33 \) km/h, express: \( 44 \) km/h.

Check: \( \frac{132}{33} – \frac{132}{44} = 4 – 3 = 1 \).

Passenger train: 33 km/h, Express train: 44 km/h

11. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Let the sides of the squares be \( x \) and \( y \), with \( x > y \).

Areas: \( x^2 + y^2 = 468 \).

Perimeters: \( 4x – 4y = 24 \implies x – y = 6 \implies x = y + 6 \).

Substitute: \( (y + 6)^2 + y^2 = 468 \implies y^2 + 12y + 36 + y^2 = 468 \implies 2y^2 + 12y – 432 = 0 \implies y^2 + 6y – 216 = 0 \).

\( a = 1 \), \( b = 6 \), \( c = -216 \).

Discriminant: \( \Delta = 6^2 – 4(1)(-216) = 36 + 864 = 900 \).

Roots: \( y = \frac{-6 \pm \sqrt{900}}{2} = \frac{-6 \pm 30}{2} \implies y = 12, -18 \). Take \( y = 12 \).

\( x = 12 + 6 = 18 \).

Check: Areas: \( 18^2 + 12^2 = 324 + 144 = 468 \), Perimeters: \( 4 \cdot 18 – 4 \cdot 12 = 72 – 48 = 24 \).

Sides: 18 m and 12 m

12. An object is thrown upwards with an initial velocity of 17 m/sec from a building with 12 m height. It is at a height of \( S = 12 + 17t – 5t^2 \) from the ground after a flight of \( t \) seconds. Find the time taken by the object to touch the ground.

Ground: \( S = 0 \), so \( 12 + 17t – 5t^2 = 0 \).

Rewrite: \( 5t^2 – 17t – 12 = 0 \).

\( a = 5 \), \( b = -17 \), \( c = -12 \).

Discriminant: \( \Delta = (-17)^2 – 4(5)(-12) = 289 + 240 = 529 \).

Roots: \( t = \frac{17 \pm \sqrt{529}}{10} = \frac{17 \pm 23}{10} \implies t = 4, -\frac{3}{5} \). Take \( t = 4 \).

Check: \( S = 12 + 17(4) – 5(4)^2 = 12 + 68 – 80 = 0 \).

Time: 4 seconds

13. If a polygon of \( n \) sides has \( \frac{1}{2} n(n-3) \) diagonals. How many sides are there in a polygon with 65 diagonals? Is there a polygon with 50 diagonals?

Set up: \( \frac{1}{2} n(n – 3) = 65 \implies n(n – 3) = 130 \implies n^2 – 3n – 130 = 0 \).

\( a = 1 \), \( b = -3 \), \( c = -130 \).

Discriminant: \( \Delta = (-3)^2 – 4(1)(-130) = 9 + 520 = 529 \).

Roots: \( n = \frac{3 \pm \sqrt{529}}{2} = \frac{3 \pm 23}{2} \implies n = 13, -10 \). Take \( n = 13 \).

Check: \( \frac{1}{2} \cdot 13 \cdot (13 – 3) = \frac{1}{2} \cdot 13 \cdot 10 = 65 \).

For 50 diagonals: \( \frac{1}{2} n(n – 3) = 50 \implies n^2 – 3n – 100 = 0 \).

Discriminant: \( \Delta = (-3)^2 – 4(1)(-100) = 9 + 400 = 409 \), not a perfect square, so no integer solutions.

Sides for 65 diagonals: 13, Polygon with 50 diagonals: No

10th Maths Quadratic Equations Exercise 5.2 Solutions

zaheerpashamd

Exercise 5.2 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on solving quadratic equations by factorisation and applying them to word problems. Mathematical expressions are rendered using MathJax.

1. Find the roots of the following quadratic equations by factorisation:

(i) \( x^2 – 3x – 10 = 0 \)

We need two numbers whose product is \( -10 \cdot 1 = -10 \) and sum is \( -3 \).

Numbers are \( -5 \) and \( 2 \): \( -5 + 2 = -3 \), \( -5 \cdot 2 = -10 \).

Rewrite: \( x^2 – 5x + 2x – 10 = 0 \).

Factor: \( x(x – 5) + 2(x – 5) = 0 \implies (x – 5)(x + 2) = 0 \).

Solve: \( x – 5 = 0 \implies x = 5 \), \( x + 2 = 0 \implies x = -2 \).

Check: \( x = 5 \): \( 5^2 – 3(5) – 10 = 25 – 15 – 10 = 0 \), \( x = -2 \): \( (-2)^2 – 3(-2) – 10 = 4 + 6 – 10 = 0 \).

Roots: \( x = 5, -2 \)

(ii) \( 2x^2 + x – 6 = 0 \)

Product: \( 2 \cdot (-6) = -12 \), sum: \( 1 \).

Numbers: \( 4 \) and \( -3 \): \( 4 + (-3) = 1 \), \( 4 \cdot (-3) = -12 \).

Rewrite: \( 2x^2 + 4x – 3x – 6 = 0 \).

Factor: \( 2x(x + 2) – 3(x + 2) = 0 \implies (x + 2)(2x – 3) = 0 \).

Solve: \( x + 2 = 0 \implies x = -2 \), \( 2x – 3 = 0 \implies x = \frac{3}{2} \).

Check: \( x = -2 \): \( 2(-2)^2 + (-2) – 6 = 8 – 2 – 6 = 0 \), \( x = \frac{3}{2} \): \( 2\left(\frac{3}{2}\right)^2 + \frac{3}{2} – 6 = 2 \cdot \frac{9}{4} + \frac{3}{2} – 6 = \frac{9}{2} + \frac{3}{2} – 6 = 0 \).

Roots: \( x = -2, \frac{3}{2} \)

(iii) \( \sqrt{2} x^2 + 7x + 5\sqrt{2} = 0 \)

Product: \( \sqrt{2} \cdot 5\sqrt{2} = 10 \), sum: \( 7 \).

Numbers: \( 5 \) and \( 2\sqrt{2} \): \( 5 + 2\sqrt{2} = 7 \), \( 5 \cdot 2\sqrt{2} = 10 \).

Rewrite: \( \sqrt{2} x^2 + 5x + 2\sqrt{2} x + 5\sqrt{2} = 0 \).

Factor: \( x(\sqrt{2} x + 5) + \sqrt{2} (\sqrt{2} x + 5) = 0 \implies (\sqrt{2} x + 5)(x + \sqrt{2}) = 0 \).

Solve: \( \sqrt{2} x + 5 = 0 \implies x = -\frac{5}{\sqrt{2}} = -\frac{5\sqrt{2}}{2} \), \( x + \sqrt{2} = 0 \implies x = -\sqrt{2} \).

Check: \( x = -\frac{5\sqrt{2}}{2} \): \( \sqrt{2} \left(-\frac{5\sqrt{2}}{2}\right)^2 + 7 \left(-\frac{5\sqrt{2}}{2}\right) + 5\sqrt{2} = \sqrt{2} \cdot \frac{25 \cdot 2}{4} – \frac{35\sqrt{2}}{2} + 5\sqrt{2} = \frac{25}{2} – \frac{30\sqrt{2}}{2} = 0 \).

Roots: \( x = -\sqrt{2}, -\frac{5\sqrt{2}}{2} \)

(iv) \( 2x^2 – x + \frac{1}{8} = 0 \)

Multiply through by 8: \( 16x^2 – 8x + 1 = 0 \).

Product: \( 16 \cdot 1 = 16 \), sum: \( -8 \).

Numbers: \( -4 \) and \( -4 \): \( -4 + (-4) = -8 \), \( -4 \cdot (-4) = 16 \).

Rewrite: \( 16x^2 – 4x – 4x + 1 = 0 \).

Factor: \( 4x(4x – 1) – 1(4x – 1) = 0 \implies (4x – 1)^2 = 0 \).

Solve: \( 4x – 1 = 0 \implies x = \frac{1}{4} \).

Check: \( 2\left(\frac{1}{4}\right)^2 – \frac{1}{4} + \frac{1}{8} = 2 \cdot \frac{1}{16} – \frac{1}{4} + \frac{1}{8} = \frac{1}{8} – \frac{1}{4} + \frac{1}{8} = 0 \).

Roots: \( x = \frac{1}{4} \) (repeated root)

(v) \( 100x^2 – 20x + 1 = 0 \)

Product: \( 100 \cdot 1 = 100 \), sum: \( -20 \).

Numbers: \( -10 \) and \( -10 \): \( -10 + (-10) = -20 \), \( -10 \cdot (-10) = 100 \).

Rewrite: \( 100x^2 – 10x – 10x + 1 = 0 \).

Factor: \( 10x(10x – 1) – 1(10x – 1) = 0 \implies (10x – 1)^2 = 0 \).

Solve: \( 10x – 1 = 0 \implies x = \frac{1}{10} \).

Check: \( 100\left(\frac{1}{10}\right)^2 – 20\left(\frac{1}{10}\right) + 1 = 100 \cdot \frac{1}{100} – 2 + 1 = 1 – 2 + 1 = 0 \).

Roots: \( x = \frac{1}{10} \) (repeated root)

(vi) \( x(x + 4) = 12 \)

Expand: \( x^2 + 4x = 12 \implies x^2 + 4x – 12 = 0 \).

Product: \( 1 \cdot (-12) = -12 \), sum: \( 4 \).

Numbers: \( 6 \) and \( -2 \): \( 6 + (-2) = 4 \), \( 6 \cdot (-2) = -12 \).

Rewrite: \( x^2 + 6x – 2x – 12 = 0 \).

Factor: \( x(x + 6) – 2(x + 6) = 0 \implies (x + 6)(x – 2) = 0 \).

Solve: \( x + 6 = 0 \implies x = -6 \), \( x – 2 = 0 \implies x = 2 \).

Check: \( x = 2 \): \( 2(2 + 4) = 12 \), \( x = -6 \): \( -6(-6 + 4) = -6(-2) = 12 \).

Roots: \( x = 2, -6 \)

(vii) \( 3x^2 – 5x + 2 = 0 \)

Product: \( 3 \cdot 2 = 6 \), sum: \( -5 \).

Numbers: \( -3 \) and \( -2 \): \( -3 + (-2) = -5 \), \( -3 \cdot (-2) = 6 \).

Rewrite: \( 3x^2 – 3x – 2x + 2 = 0 \).

Factor: \( 3x(x – 1) – 2(x – 1) = 0 \implies (x – 1)(3x – 2) = 0 \).

Solve: \( x – 1 = 0 \implies x = 1 \), \( 3x – 2 = 0 \implies x = \frac{2}{3} \).

Check: \( x = 1 \): \( 3(1)^2 – 5(1) + 2 = 3 – 5 + 2 = 0 \), \( x = \frac{2}{3} \): \( 3\left(\frac{2}{3}\right)^2 – 5\left(\frac{2}{3}\right) + 2 = 3 \cdot \frac{4}{9} – \frac{10}{3} + 2 = \frac{4}{3} – \frac{10}{3} + 2 = 0 \).

Roots: \( x = 1, \frac{2}{3} \)

(viii) \( x – \frac{3}{x} = 2 \) (\( x \neq 0 \))

Multiply through by \( x \): \( x^2 – 3 = 2x \implies x^2 – 2x – 3 = 0 \).

Product: \( 1 \cdot (-3) = -3 \), sum: \( -2 \).

Numbers: \( -3 \) and \( 1 \): \( -3 + 1 = -2 \), \( -3 \cdot 1 = -3 \).

Rewrite: \( x^2 – 3x + x – 3 = 0 \).

Factor: \( x(x – 3) + 1(x – 3) = 0 \implies (x – 3)(x + 1) = 0 \).

Solve: \( x – 3 = 0 \implies x = 3 \), \( x + 1 = 0 \implies x = -1 \).

Check: \( x = 3 \): \( 3 – \frac{3}{3} = 2 \), \( x = -1 \): \( -1 – \frac{3}{-1} = -1 + 3 = 2 \).

Roots: \( x = 3, -1 \)

(ix) \( 3(x – 4)^2 – 5(x – 4) = 12 \)

Substitute \( u = x – 4 \). Then: \( 3u^2 – 5u = 12 \implies 3u^2 – 5u – 12 = 0 \).

Product: \( 3 \cdot (-12) = -36 \), sum: \( -5 \).

Numbers: \( -9 \) and \( 4 \): \( -9 + 4 = -5 \), \( -9 \cdot 4 = -36 \).

Rewrite: \( 3u^2 – 9u + 4u – 12 = 0 \).

Factor: \( 3u(u – 3) + 4(u – 3) = 0 \implies (u – 3)(3u + 4) = 0 \).

Solve: \( u – 3 = 0 \implies u = 3 \), \( 3u + 4 = 0 \implies u = -\frac{4}{3} \).

Back-substitute: \( x – 4 = 3 \implies x = 7 \), \( x – 4 = -\frac{4}{3} \implies x = -\frac{4}{3} + 4 = \frac{8}{3} \).

Check: \( x = 7 \): \( 3(7 – 4)^2 – 5(7 – 4) = 3(3)^2 – 5(3) = 27 – 15 = 12 \).

Roots: \( x = 7, \frac{8}{3} \)

2. Find two numbers whose sum is 27 and product is 182.

Let the numbers be \( x \) and \( 27 – x \).

Product: \( x(27 – x) = 182 \implies 27x – x^2 = 182 \implies x^2 – 27x + 182 = 0 \).

Product: \( 1 \cdot 182 = 182 \), sum: \( -27 \).

Numbers: \( -14 \) and \( -13 \): \( -14 + (-13) = -27 \), \( -14 \cdot (-13) = 182 \).

Rewrite: \( x^2 – 14x – 13x + 182 = 0 \).

Factor: \( x(x – 14) – 13(x – 14) = 0 \implies (x – 14)(x – 13) = 0 \).

Solve: \( x = 14 \), \( x = 13 \).

Numbers: \( 14 \) and \( 27 – 14 = 13 \).

Check: Sum: \( 14 + 13 = 27 \), Product: \( 14 \cdot 13 = 182 \).

Numbers: 14 and 13

3. Find two consecutive positive integers, sum of whose squares is 613.

Let the integers be \( x \) and \( x + 1 \).

Sum of squares: \( x^2 + (x + 1)^2 = 613 \).

Expand: \( x^2 + x^2 + 2x + 1 = 613 \implies 2x^2 + 2x + 1 – 613 = 0 \implies 2x^2 + 2x – 612 = 0 \implies x^2 + x – 306 = 0 \).

Product: \( -306 \), sum: \( 1 \).

Numbers: \( 18 \) and \( -17 \): \( 18 + (-17) = 1 \), \( 18 \cdot (-17) = -306 \).

Rewrite: \( x^2 + 18x – 17x – 306 = 0 \).

Factor: \( x(x + 18) – 17(x + 18) = 0 \implies (x + 18)(x – 17) = 0 \).

Solve: \( x = -18 \), \( x = 17 \). Since positive integers are required, take \( x = 17 \).

Integers: \( 17 \) and \( 18 \).

Check: \( 17^2 + 18^2 = 289 + 324 = 613 \).

Integers: 17 and 18

4. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Let the base be \( x \) cm, altitude = \( x – 7 \).

By Pythagoras: \( x^2 + (x – 7)^2 = 13^2 \).

Expand: \( x^2 + x^2 – 14x + 49 = 169 \implies 2x^2 – 14x + 49 – 169 = 0 \implies 2x^2 – 14x – 120 = 0 \implies x^2 – 7x – 60 = 0 \).

Product: \( -60 \), sum: \( -7 \).

Numbers: \( -12 \) and \( 5 \): \( -12 + 5 = -7 \), \( -12 \cdot 5 = -60 \).

Rewrite: \( x^2 – 12x + 5x – 60 = 0 \).

Factor: \( x(x – 12) + 5(x – 12) = 0 \implies (x – 12)(x + 5) = 0 \).

Solve: \( x = 12 \), \( x = -5 \). Since length is positive, \( x = 12 \).

Base: \( 12 \) cm, altitude: \( 12 – 7 = 5 \) cm.

Check: \( 12^2 + 5^2 = 144 + 25 = 169 = 13^2 \).

Sides: Base = 12 cm, Altitude = 5 cm

5. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.

Let the number of articles be \( x \).

Cost of each article: \( 2x + 3 \).

Total cost: \( x (2x + 3) = 90 \implies 2x^2 + 3x – 90 = 0 \).

Product: \( 2 \cdot (-90) = -180 \), sum: \( 3 \).

Numbers: \( 15 \) and \( -12 \): \( 15 + (-12) = 3 \), \( 15 \cdot (-12) = -180 \).

Rewrite: \( 2x^2 + 15x – 12x – 90 = 0 \).

Factor: \( x(2x + 15) – 6(2x + 15) = 0 \implies (2x + 15)(x – 6) = 0 \).

Solve: \( 2x + 15 = 0 \implies x = -\frac{15}{2} \), \( x – 6 = 0 \implies x = 6 \). Take \( x = 6 \).

Number of articles: \( 6 \), cost per article: \( 2(6) + 3 = 15 \).

Check: Total cost: \( 6 \cdot 15 = 90 \).

Articles: 6, Cost per article: Rs 15

6. Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.

Let length be \( x \) m, breadth be \( y \) m.

Perimeter: \( 2(x + y) = 28 \implies x + y = 14 \implies y = 14 – x \).

Area: \( x y = 40 \).

Substitute: \( x (14 – x) = 40 \implies 14x – x^2 = 40 \implies x^2 – 14x + 40 = 0 \).

Product: \( 40 \), sum: \( -14 \).

Numbers: \( -10 \) and \( -4 \): \( -10 + (-4) = -14 \), \( -10 \cdot (-4) = 40 \).

Rewrite: \( x^2 – 10x – 4x + 40 = 0 \).

Factor: \( x(x – 10) – 4(x – 10) = 0 \implies (x – 10)(x – 4) = 0 \).

Solve: \( x = 10 \), \( x = 4 \).

If \( x = 10 \), \( y = 14 – 10 = 4 \). If \( x = 4 \), \( y = 10 \).

Check: Perimeter: \( 2(10 + 4) = 28 \), Area: \( 10 \cdot 4 = 40 \).

Dimensions: 10 m and 4 m

7. The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm then find its base and altitude.

Let the altitude be \( x \) cm, base = \( x + 4 \).

Area: \( \frac{1}{2} \times \text{base} \times \text{altitude} = 48 \implies \frac{1}{2} (x + 4) x = 48 \implies x(x + 4) = 96 \).

\( x^2 + 4x – 96 = 0 \).

Product: \( -96 \), sum: \( 4 \).

Numbers: \( 12 \) and \( -8 \): \( 12 + (-8) = 4 \), \( 12 \cdot (-8) = -96 \).

Rewrite: \( x^2 + 12x – 8x – 96 = 0 \).

Factor: \( x(x + 12) – 8(x + 12) = 0 \implies (x + 12)(x – 8) = 0 \).

Solve: \( x = -12 \), \( x = 8 \). Take \( x = 8 \).

Altitude: \( 8 \) cm, base: \( 8 + 4 = 12 \) cm.

Check: Area: \( \frac{1}{2} \times 12 \times 8 = 48 \).

Base: 12 cm, Altitude: 8 cm

8. Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km apart, find the average speed of each train.

Let the speed of the second train (north) be \( x \) km/h, first train (west) = \( x + 5 \) km/h.

After 2 hours: Distance west = \( 2(x + 5) \), distance north = \( 2x \).

Distance apart (right triangle): \( \sqrt{(2x)^2 + (2(x + 5))^2} = 50 \).

Simplify: \( 4x^2 + 4(x^2 + 10x + 25) = 2500 \implies 4x^2 + 4x^2 + 40x + 100 = 2500 \implies 8x^2 + 40x – 2400 = 0 \implies x^2 + 5x – 300 = 0 \).

Product: \( -300 \), sum: \( 5 \).

Numbers: \( 20 \) and \( -15 \): \( 20 + (-15) = 5 \), \( 20 \cdot (-15) = -300 \).

Rewrite: \( x^2 + 20x – 15x – 300 = 0 \).

Factor: \( x(x + 20) – 15(x + 20) = 0 \implies (x + 20)(x – 15) = 0 \).

Solve: \( x = -20 \), \( x = 15 \). Take \( x = 15 \).

Second train: \( 15 \) km/h, first train: \( 15 + 5 = 20 \) km/h.

Check: Distances: \( 2 \cdot 15 = 30 \), \( 2 \cdot 20 = 40 \). \( \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = 50 \).

Speeds: First train (west): 20 km/h, Second train (north): 15 km/h

9. In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was Rs 1600, how many boys were there in the class?

Let the number of boys be \( x \), girls = \( 60 – x \).

Boys contribute: \( x \cdot (60 – x) \), girls contribute: \( (60 – x) \cdot x \).

Total: \( x(60 – x) + (60 – x)x = 2x(60 – x) = 1600 \).

Simplify: \( 2x(60 – x) = 1600 \implies 120x – 2x^2 = 1600 \implies x^2 – 60x + 800 = 0 \).

Product: \( 800 \), sum: \( -60 \).

Numbers: \( -40 \) and \( -20 \): \( -40 + (-20) = -60 \), \( -40 \cdot (-20) = 800 \).

Rewrite: \( x^2 – 40x – 20x + 800 = 0 \).

Factor: \( x(x – 40) – 20(x – 40) = 0 \implies (x – 40)(x – 20) = 0 \).

Solve: \( x = 40 \), \( x = 20 \).

If \( x = 40 \), girls = \( 20 \). If \( x = 20 \), girls = \( 40 \). Both are symmetric.

Check: \( 40 \cdot 20 + 20 \cdot 40 = 800 + 800 = 1600 \).

Number of boys: 40 or 20 (both possible, but typically \( 40 \) is chosen as “boys”)

10. A motor boat heads upstream a distance of 24 km in a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed in still water?

Let the boat’s speed in still water be \( x \) km/h.

Upstream speed: \( x – 3 \), downstream speed: \( x + 3 \).

Time upstream: \( \frac{24}{x-3} \), time downstream: \( \frac{24}{x+3} \).

Total time: \( \frac{24}{x-3} + \frac{24}{x+3} = 6 \).

Simplify: \( 24 \left( \frac{(x+3) + (x-3)}{(x-3)(x+3)} \right) = 6 \implies 24 \cdot \frac{2x}{x^2 – 9} = 6 \implies 48x = 6(x^2 – 9) \implies 8x = x^2 – 9 \implies x^2 – 8x – 9 = 0 \).

Product: \( -9 \), sum: \( -8 \).

Numbers: \( -9 \) and \( 1 \): \( -9 + 1 = -8 \), \( -9 \cdot 1 = -9 \).

Rewrite: \( x^2 – 9x + x – 9 = 0 \).

Factor: \( x(x – 9) + 1(x – 9) = 0 \implies (x – 9)(x + 1) = 0 \).

Solve: \( x = 9 \), \( x = -1 \). Take \( x = 9 \).

Check: \( \frac{24}{9-3} + \frac{24}{9+3} = \frac{24}{6} + \frac{24}{12} = 4 + 2 = 6 \).

Speed in still water: 9 km/h

10th Maths Quadratic Equations Exercise 5.1 Solutions

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Exercise 5.1 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on quadratic equations. Mathematical expressions are rendered using MathJax.

1. Check whether the following are quadratic equations:

(i) \( (x + 1)^2 = 2(x – 3) \)

Expand the equation: \( (x + 1)^2 = x^2 + 2x + 1 \).

So, \( x^2 + 2x + 1 = 2(x – 3) = 2x – 6 \).

Simplify: \( x^2 + 2x + 1 – 2x + 6 = 0 \implies x^2 + 7 = 0 \).

This is a quadratic equation because it has the form \( ax^2 + bx + c = 0 \) with \( a = 1 \), \( b = 0 \), \( c = 7 \).

Conclusion: Yes, it is a quadratic equation.

(ii) \( x^2 – 2x = (-2)(3 – x) \)

Expand the right side: \( (-2)(3 – x) = -6 + 2x \).

So, \( x^2 – 2x = -6 + 2x \).

Simplify: \( x^2 – 2x – 2x + 6 = 0 \implies x^2 – 4x + 6 = 0 \).

This is a quadratic equation (\( a = 1 \), \( b = -4 \), \( c = 6 \)).

Conclusion: Yes, it is a quadratic equation.

(iii) \( (x – 2)(x + 1) = (x – 1)(x + 3) \)

Expand both sides: Left: \( (x – 2)(x + 1) = x^2 + x – 2x – 2 = x^2 – x – 2 \).

Right: \( (x – 1)(x + 3) = x^2 + 3x – x – 3 = x^2 + 2x – 3 \).

So, \( x^2 – x – 2 = x^2 + 2x – 3 \).

Simplify: \( x^2 – x – 2 – x^2 – 2x + 3 = 0 \implies -3x + 1 = 0 \).

This is a linear equation, not a quadratic equation (no \( x^2 \) term).

Conclusion: No, it is not a quadratic equation.

(iv) \( (x – 3)(2x + 1) = x(x + 5) \)

Expand: Left: \( (x – 3)(2x + 1) = 2x^2 + x – 6x – 3 = 2x^2 – 5x – 3 \).

Right: \( x(x + 5) = x^2 + 5x \).

So, \( 2x^2 – 5x – 3 = x^2 + 5x \).

Simplify: \( 2x^2 – 5x – 3 – x^2 – 5x = 0 \implies x^2 – 10x – 3 = 0 \).

This is a quadratic equation (\( a = 1 \), \( b = -10 \), \( c = -3 \)).

Conclusion: Yes, it is a quadratic equation.

(v) \( (2x – 1)(x – 3) = (x + 5)(x – 1) \)

Expand: Left: \( (2x – 1)(x – 3) = 2x^2 – 6x – x + 3 = 2x^2 – 7x + 3 \).

Right: \( (x + 5)(x – 1) = x^2 – x + 5x – 5 = x^2 + 4x – 5 \).

So, \( 2x^2 – 7x + 3 = x^2 + 4x – 5 \).

Simplify: \( 2x^2 – 7x + 3 – x^2 – 4x + 5 = 0 \implies x^2 – 11x + 8 = 0 \).

This is a quadratic equation (\( a = 1 \), \( b = -11 \), \( c = 8 \)).

Conclusion: Yes, it is a quadratic equation.

(vi) \( x^2 + 3x + 1 = (x – 2)^2 \)

Expand: Right: \( (x – 2)^2 = x^2 – 4x + 4 \).

So, \( x^2 + 3x + 1 = x^2 – 4x + 4 \).

Simplify: \( x^2 + 3x + 1 – x^2 + 4x – 4 = 0 \implies 7x – 3 = 0 \).

This is a linear equation, not a quadratic equation.

Conclusion: No, it is not a quadratic equation.

(vii) \( (x + 2)^3 = 2x(x^2 – 1) \)

Expand: Left: \( (x + 2)^3 = x^3 + 3x^2(2) + 3x(2^2) + 2^3 = x^3 + 6x^2 + 12x + 8 \).

Right: \( 2x(x^2 – 1) = 2x^3 – 2x \).

So, \( x^3 + 6x^2 + 12x + 8 = 2x^3 – 2x \).

Simplify: \( x^3 + 6x^2 + 12x + 8 – 2x^3 + 2x = 0 \implies -x^3 + 6x^2 + 14x + 8 = 0 \).

This is a cubic equation (highest degree 3), not a quadratic equation.

Conclusion: No, it is not a quadratic equation.

(viii) \( x^3 – 4x^2 – x + 1 = (x – 2)^3 \)

Expand: Right: \( (x – 2)^3 = x^3 – 3x^2(2) + 3x(2^2) – 2^3 = x^3 – 6x^2 + 12x – 8 \).

So, \( x^3 – 4x^2 – x + 1 = x^3 – 6x^2 + 12x – 8 \).

Simplify: \( x^3 – 4x^2 – x + 1 – x^3 + 6x^2 – 12x + 8 = 0 \implies 2x^2 – 13x + 9 = 0 \).

This is a quadratic equation (\( a = 2 \), \( b = -13 \), \( c = 9 \)).

Conclusion: Yes, it is a quadratic equation.

2. Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m². The length of the plot is one meter more than twice its breadth. We need to find the length and breadth of the plot.

Let the breadth of the plot be \( x \) meters.

Length is one meter more than twice the breadth: \( 2x + 1 \).

Area of the rectangle: \( \text{length} \times \text{breadth} = 528 \).

So, \( x (2x + 1) = 528 \).

Expand: \( 2x^2 + x = 528 \implies 2x^2 + x – 528 = 0 \).

Quadratic equation: \( 2x^2 + x – 528 = 0 \)

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Let the first integer be \( x \).

The next consecutive integer is \( x + 1 \).

Their product is 306: \( x (x + 1) = 306 \).

Expand: \( x^2 + x = 306 \implies x^2 + x – 306 = 0 \).

Quadratic equation: \( x^2 + x – 306 = 0 \)

(iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.

Let Rohan’s present age be \( x \) years.

His mother’s present age is \( x + 26 \).

After 3 years: Rohan’s age = \( x + 3 \), mother’s age = \( x + 29 \).

Product of their ages: \( (x + 3)(x + 29) = 360 \).

Expand: \( x^2 + 29x + 3x + 87 = 360 \implies x^2 + 32x + 87 – 360 = 0 \implies x^2 + 32x – 273 = 0 \).

Quadratic equation: \( x^2 + 32x – 273 = 0 \)

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Let the speed of the train be \( x \) km/h.

Time to travel 480 km at speed \( x \): \( \frac{480}{x} \) hours.

If speed is \( x – 8 \), time taken: \( \frac{480}{x-8} \).

The slower speed takes 3 hours more: \( \frac{480}{x-8} = \frac{480}{x} + 3 \).

Simplify: \( \frac{480}{x-8} – \frac{480}{x} = 3 \implies 480 \left( \frac{x – (x-8)}{x(x-8)} \right) = 3 \implies \frac{480 \cdot 8}{x(x-8)} = 3 \implies \frac{3840}{x^2 – 8x} = 3 \).

\( 3840 = 3(x^2 – 8x) \implies 1280 = x^2 – 8x \implies x^2 – 8x – 1280 = 0 \).

Quadratic equation: \( x^2 – 8x – 1280 = 0 \)

QUADRATIC EQUATIONS

zaheerpashamd

QUADRATIC EQUATIONS Key Points

Standard form of quadratic equation in variable 𝑥 is 𝒂𝒙^𝟐 + 𝒃𝒙 + 𝒄 = 𝟎, where 𝑎, 𝑏, 𝑐 are
real numbers and 𝑎 ≠ 0.
y = 𝑎𝑥²+ 𝑏𝑥 + 𝑐 is called a quadratic function.
Uses of Quadratic functions.
(i) When the rocket is fired upward, then the path of the rocket is defined by a
‘quadratic function.’
(ii) Shapes of the satellite dish, reflecting mirror in a telescope, lens of the eye glasses
and orbits of the celestial objects are defined by the quadratic equations.
(iii) The path of a projectile is defined by quadratic function.
(iv) When the breaks are applied to a vehicle, the stopping distance is calculated by using
quadratic equation
Roots of quadratic equation: A real number 𝛼 is called a root of the quadratic equation
𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0, if 𝑎𝛼² + 𝑏𝛼 + 𝑐 = 0
Quadratic formula: The roots of a quadratic equation 𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0 are given by

Nature of roots:
The nature of roots of a quadratic equation 𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0 depends on 𝑏² − 4𝑎𝑐 is
called the discriminant
(i) 𝐼𝑓 𝒃^𝟐 − 𝟒𝒂𝒄 > 𝟎 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑎𝑟𝑒 𝑟𝑒𝑎𝑙 𝑎𝑛𝑑 𝑑𝑖𝑠𝑡𝑖𝑛𝑐𝑡
(ii) 𝐼𝑓 𝒃^𝟐 − 𝟒𝒂𝒄 = 𝟎 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑎𝑟𝑒 𝑟𝑒𝑎𝑙 𝑎𝑛𝑑 𝑒𝑞𝑢𝑎𝑙
(iii) 𝐼𝑓 𝒃^𝟐 − 𝟒𝒂𝒄 < 𝟎 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑟𝑒𝑎𝑙

Try This

Check whether the following equations are quadratic or not ?
(i) 𝑥 ²– 6𝑥 – 4 = 0
Sol: Quadratic equation

(ii) 𝑥 ³– 6𝑥 ² + 2𝑥 -1 = 0
Sol: Not a quadratic equation(Cubic equation)

(iii) 7x = 2x ²
Sol: 2𝑥² − 7𝑥 = 0
Quadratic equation

𝑣) (2𝑥 + 1) (3𝑥 + 1) = 𝑏(𝑥 -1) (𝑥 -2)
𝑆𝑜𝑙: (2𝑥 + 1) (3𝑥 + 1) = 𝑏(𝑥 -1) (𝑥 -2)
2𝑥(3𝑥 + 1) + 1(3𝑥 + 1) = 𝑏[𝑥(𝑥 − 2) − 1(𝑥 − 2)]
6𝑥² + 2𝑥 + 3𝑥 + 1 = 𝑏[𝑥² − 2𝑥 − 𝑥 + 2]
6𝑥² + 5𝑥 + 1 = 𝑏[𝑥² − 3𝑥 + 2]
6𝑥² + 5𝑥 + 1 = 𝑏𝑥² − 3𝑏𝑥 + 2𝑏
6𝑥² − 𝑏𝑥² + 5𝑥 + 3𝑏𝑥 + 1 − 2𝑏 = 0
(6 − 𝑏)𝑥² + (5 + 3𝑏)𝑥 + 1 − 2𝑏 = 0
This is a quadratic equation.

(vi) 3y² = 192
Sol: 3y² – 192=0
This is a quadratic equation

Example-1.

i. Raju and Rajendar together have 45 marbles. Both of them lost 5 marbles each, and the
product of the number of marbles now they have is 124. We would like to find out how
many marbles they had previously. Represent the situation mathematically.
Sol: Total marbles=45
Let the number of marbles at Raju = 𝑥
Then the number of marbles at Rajendar = 45 − 𝑥
If both of them lost 5 marbles each then
The number of marbles at Raju = 𝑥 − 5
The number of marbles at Rajendar = 45 − 𝑥 − 5 = 40 − 𝑥
The product of remaining marbles=124
⇒ (𝑥 − 5)(40 − 𝑥) = 124
⇒ 40𝑥 − 𝑥² − 200 + 5𝑥 = 124
⇒ −𝑥² + 45𝑥 − 200 − 124 = 0
⇒ −𝑥² + 45𝑥 − 324 = 0
⇒ 𝑥² − 45𝑥 + 324 = 0 (multiply with −1)

ii. The hypotenuse of a right triangle is 25 cm. We know that the difference in lengths of the
other two sides is 5 cm. We would like to find out the length of the two sides?

Sol: Let the length of smaller side = 𝑥 𝑐𝑚
The length of larger side = (𝑥 + 5)𝑐𝑚
Length of hypotenuse = 25 𝑐𝑚
In a right angle triangle we know that
(𝑠𝑖𝑑𝑒)² + (𝑠𝑖𝑑𝑒)² = (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒)²
(𝑥)² + (𝑥 + 5)² = (25)²
𝑥² + 𝑥² + 2 × 𝑥 × 5 + 52 = 625
2𝑥² + 10𝑥 + 25 − 625 = 0
2𝑥² + 10𝑥 + −600 = 0
𝑥² + 5𝑥 − 300 = 0
Required quadratic equation: 𝑥² + 5𝑥 − 300

Example-2. Check whether the following are quadratic equations:
i. (𝑥 − 2)² + 1 = 2𝑥 − 3
Sol: (𝑥 − 2)² + 1 = 2𝑥 − 3
⇒ 𝑥² − 4𝑥 + 4 + 1 − 2𝑥 + 3 = 0
⇒ 𝑥² − 6𝑥 + 8 = 0
The given equation is a quadratic equation.

ii. 𝑥(𝑥 + 1) + 8 = (𝑥 + 2)(𝑥 − 2)
Sol: 𝑥(𝑥 + 1) + 8 = (𝑥 + 2)(𝑥 − 2)
⇒ 𝑥² + 𝑥 + 8 = 𝑥² − 22
⇒ 𝑥² + 𝑥 + 8 − 𝑥² + 4 = 0
⇒ 𝑥 + 12 = 0
The given equation is not a quadratic equation.

iii. 𝑥(2𝑥 + 3) = 𝑥² + 1
Sol: 𝑥(2𝑥 + 3) = 𝑥² + 1
⇒ 2𝑥² + 3𝑥 − 𝑥² − 1 = 0
⇒ 𝑥² + 3𝑥 − 1 = 0
The given equation is a quadratic equation

iv. (𝑥 + 2)³ = 𝑥³ − 4
Sol: (𝑥 + 2)³ = 𝑥³ − 4 (𝑎 + 𝑏)³ = 𝑎³ + 𝑏³ + 3𝑎𝑏(𝑎 + 𝑏)
⇒ 𝑥³ + 23 + 3 × 𝑥 × 2(𝑥 + 2) = 𝑥³ − 4
⇒ 𝑥³ + 8 + 6𝑥(𝑥 + 2) = 𝑥³ − 4 ⇒ 𝑥³ + 8 + 6𝑥² + 12𝑥 − 𝑥³ + 4 = 0
⇒ 6𝑥² + 12x + 12 = 0
The given equation is a quadratic equation

EXERCISE-5.1

Check whether the following are quadratic equations:
i. (𝑥 + 1)² = 2(𝑥 − 3)
Sol: (𝑥 + 1)² = 2(𝑥 − 3)
⇒ 𝑥² + 2𝑥 + 1 = 2𝑥 − 6
⇒ 𝑥² + 2𝑥 + 1 − 2𝑥 + 6 = 0
⇒ 𝑥² + 7 = 0
The given equation is a quadratic equation.
ii. 𝑥² − 2𝑥 = (−2)(3 − 𝑥)
Sol: 𝑥² − 2𝑥 = (−2)(3 − 𝑥)
⇒ 𝑥² − 2𝑥 = −6 + 2𝑥
⇒ 𝑥² − 2𝑥 + 6 − 2𝑥 = 0
⇒ 𝑥² − 4𝑥 + 6 = 0
The given equation is a quadratic equation.
iii. (𝑥 – 2)(𝑥 + 1) = (𝑥 – 1)(𝑥 + 3)
Sol: (𝑥 – 2)(𝑥 + 1) = (𝑥 – 1)(𝑥 + 3)
⇒ 𝑥² + 𝑥 − 2𝑥 − 2 = 𝑥² + 3𝑥 − 𝑥 − 3
⇒ 𝑥² − 𝑥 − 2 = 𝑥² + 2𝑥 − 3
⇒ 𝑥² − 𝑥 − 2 − 𝑥²− 2𝑥 + 3 = 0
⇒ −3𝑥 + 1 = 0
The given equation is not a quadratic equation.
iv. (𝑥 – 3)(2𝑥 + 1) = 𝑥(𝑥 + 5)
Sol: (𝑥 – 3)(2𝑥 + 1) = 𝑥(𝑥 + 5)
⇒ 2𝑥² + 𝑥 − 6𝑥 − 3 = 𝑥² + 5𝑥
⇒ 2𝑥² − 5𝑥 − 3 − 𝑥² − 5𝑥 = 0
⇒ 𝑥² − 10𝑥 − 3 = 0
The given equation is a quadratic equation.
v. (2𝑥 – 1)(𝑥 – 3) = (𝑥 + 5)(𝑥 – 1)
Sol: (2𝑥 – 1)(𝑥 – 3) = (𝑥 + 5)(𝑥 – 1)
⇒ 2𝑥² − 6𝑥 − 𝑥 + 3 = 𝑥2 − 𝑥 + 5𝑥 − 5
⇒ 2𝑥² − 7𝑥 + 3 = 𝑥² + 4𝑥 − 5
⇒ 2𝑥² − 7𝑥 + 3 − 𝑥² − 4𝑥 + 5 = 0 ⇒ 𝑥² − 11𝑥 + 8 = 0 The given equation is a quadratic equation.

vi. 𝑥² + 3𝑥 + 1 = (𝑥 – 2)²
Sol: 𝑥² + 3𝑥 + 1 = (𝑥 – 2)²
⇒ 𝑥² + 3𝑥 + 1 = 𝑥² − 4𝑥 + 4
⇒ 𝑥²+ 3𝑥 + 1 − 𝑥² + 4𝑥 − 4 = 0
⇒ 7𝑥 − 3 = 0
The given equation is not a quadratic equation.
vii. (𝑥 + 2)³ = 2𝑥(𝑥² − 1)
Sol: (𝑥 + 2)³ = 2𝑥(𝑥² − 1)
⇒ 𝑥³ + 2³ + 3 × 𝑥 × 2(𝑥 + 2) = 2𝑥³ − 2𝑥
⇒ 𝑥³ + 8 + 6𝑥(𝑥 + 2) = 2𝑥³ − 2𝑥
⇒ 𝑥³ + 8 + 6𝑥² + 12𝑥 − 2𝑥³ + 2𝑥 = 0
⇒ −𝑥³ + 6𝑥² + 14𝑥 + 8 = 0
The given equation is not a quadratic equation.
viii. 𝑥³ − 4𝑥² − 𝑥 + 1 = (𝑥 − 2)³
Sol: 𝑥³ − 4𝑥² − 𝑥 + 1 = (𝑥 − 2)³
⇒ 𝑥³ − 4𝑥² − 𝑥 + 1 = 𝑥³ − 2³ − 3 × 𝑥 × 2(𝑥 − 2)
⇒ 𝑥³ − 4𝑥² − 𝑥 + 1 = 𝑥³ − 8 − 6𝑥² + 12𝑥
⇒ 𝑥³ − 4𝑥² − 𝑥 + 1 − 𝑥³ + 8 + 6𝑥² − 12𝑥 = 0
⇒ 2𝑥² − 13𝑥 + 9 = 0
The given equation is a quadratic equation.

i. The area of a rectangular plot is 528 m² . The length of the plot (in metres) is one more
than twice its breadth. We need to find the length and breadth of the plot.
Sol: Let breadth of rectangular plot (𝑏) = 𝑥 𝑚
Length of rectangular plot (𝑙) = (2𝑥 + 1)𝑚
Given area of the rectangular plot = 528 𝑚²
𝑙 × 𝑏 = 528
(2𝑥 + 1) × 𝑥 = 528
2𝑥² + 𝑥 − 528 = 0
This is the required quadratic equation.
ii. The product of two consecutive positive integers is 306. We need to find the integers.
Sol: Let the two consecutive positive integers be 𝑥 , 𝑥 + 1
Given the product of two consecutive positive integers = 306
𝑥 × (𝑥 + 1) = 306 ⟹ 𝑥² + 𝑥 − 306 = 0
This is the required quadratic equation
iii. Rohan’s mother is 26 years older than him. The product of their ages after 3 years will
be 360 years. We need to find Rohan’s present age.
Sol: Let Rohan’s age=𝑥 years
Rohan’s mother age=(𝑥 + 26)𝑦𝑒𝑎𝑟𝑠
After 3 years
Rohan’s age=𝑥 + 3 years
Rohan’s mother age=(𝑥 + 26 + 3) = (𝑥 + 29)𝑦𝑒𝑎𝑟𝑠
Given the product of their ages after 3 years=360 years
⇒ (𝑥 + 3)(𝑥 + 29) = 360
⇒ 𝑥² + 29𝑥 + 3𝑥 + 87 − 360 = 0
⇒ 𝑥² + 32𝑥 − 273 = 0
This is the required quadratic equation.

iv. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h
less, then it would have taken 3 hours more to cover the same distance. We need to
find the speed of the train.
Sol: Let the speed of the train= 𝑥 𝑘𝑚/ℎ
Distance= 480 km
𝑇𝑖𝑚𝑒(𝑇1) =𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑆𝑝𝑒𝑒𝑑=480-𝑥/ℎ
If the speed had been 8 km/h less then speed=(𝑥 − 8) 𝑘𝑚/ℎ
𝑇𝑖𝑚𝑒(𝑇2) =𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑆𝑝𝑒𝑒𝑑=480/𝑥 − 8*ℎ
Difference of times(𝑇2 − 𝑇1) =3h

SOLUTIONS OF A QUADRATIC EQUATIONS BY FACTORISATION

A real number  is called a root of the quadratic equation 𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0, if 𝑎𝛼² + 𝑏 𝑎 + 𝑐 = 0
We also say that x = 𝛼 is a solution of the quadratic equation.

Example-3. Find the roots of the equation 2𝑥² − 5𝑥 + 3 = 0, by factorisation.
Sol: 2𝑥² − 5𝑥 + 3 = 0
2𝑥² − 2𝑥 − 3𝑥 + 3 = 0
2𝑥(𝑥 − 1) − 3(𝑥 − 1) = 0
(𝑥 − 1)(2𝑥 − 3) = 0
𝑥 − 1 = 0 𝑜𝑟 2𝑥 − 3 = 0
𝑥 = 1 𝑜𝑟 𝑥 =3/2
∴ 1 𝑎𝑛𝑑 3/2 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2𝑥² − 5𝑥 + 3 = 0

EXERCISE-5.2

Find the roots of the following quadratic equations by factorisation.
(i) 𝑥² − 3𝑥 − 10 = 0
Sol: 𝑥² − 3𝑥 − 10 = 0
𝑥² − 2𝑥 + 5𝑥 − 10 = 0
𝑥(𝑥 − 2) + 5(𝑥 − 2) = 0
(𝑥 − 2)(𝑥 + 5) = 0
𝑥 − 2 = 0 𝑜𝑟 𝑥 + 5 = 0
𝑥 = 2 𝑜𝑟 𝑥 = −5
The roots of 𝑥² − 3𝑥 − 10 = 0 are 2 𝑎𝑛𝑑 − 5
(ii) 2𝑥² + 𝑥 − 6 = 0
Sol: 2𝑥² + 𝑥 − 6 = 0
2𝑥² − 3𝑥 + 4𝑥 − 6 = 0
𝑥(2𝑥 − 3) + 2(2𝑥 − 3) = 0
(2𝑥 − 3)(𝑥 + 2) = 0
2𝑥 − 3 = 0 𝑜𝑟 𝑥 + 2 = 0
𝑥 =3/2 𝑜𝑟 𝑥 = −2
The roots of 2𝑥² + 𝑥 − 6 = 0 𝑎𝑟𝑒 3/2 𝑎𝑛𝑑 − 2
(iii) √2𝑥² + 7𝑥 + 5√2 = 0
Sol: √2𝑥² + 7𝑥 + 5√2 = 0
√2𝑥² + 2𝑥 + 5𝑥 + 5√2 = 0
√2𝑥(𝑥 + √2) + 5(𝑥 + √2) = 0
(𝑥 + √2)(√2𝑥 + 5) = 0

𝑥 + √2 = 0 𝑜𝑟 √2𝑥 + 5 = 0
𝑥 = −√2 𝑜𝑟 𝑥 =−5/√2
𝑇ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 √2𝑥² + 7𝑥 + 5√2 = 0 𝑎𝑟𝑒 − √2 𝑎𝑛𝑑 −5/√2

(iv) 2𝑥² − 𝑥 + 1/8= 0
Sol: 2𝑥² − 𝑥 + 1/8= 0
Multiply with ‘8’
8 × 2𝑥² − 8 × 𝑥 + 8 ×1/8= 8 × 0
16𝑥² − 8𝑥 + 1 = 0
16𝑥² − 4𝑥 − 4𝑥 + 1 = 0
4𝑥(4𝑥 − 1) − 1(4𝑥 − 1) = 0
(4𝑥 − 1)(4𝑥 − 1) = 0
4𝑥 − 1 = 0 𝑜𝑟 4𝑥 − 1 = 0
𝑥 =1/4 𝑜𝑟 𝑥 =1/4
𝑇ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 2𝑥² − 𝑥 + 1/8= 0 𝑎𝑟𝑒 1/4 𝑎𝑛𝑑 1/4
.
(v) 100𝑥² − 20𝑥 + 1 = 0
Sol: 100𝑥² − 20𝑥 + 1 = 0
100𝑥² − 10𝑥 − 10𝑥 + 1 = 0
10𝑥(10𝑥 − 1) − 1(10𝑥 − 1) = 0
(10𝑥 − 1)(10𝑥 − 1) = 0
10𝑥 − 1 = 0 𝑜𝑟 10𝑥 − 1 = 0
𝑥 =1/10 𝑜𝑟 𝑥 =1/10
𝑇ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 100𝑥² − 20𝑥 + 1 = 0 𝑎𝑟𝑒 1/10 𝑎𝑛𝑑 1/10
(vi) 𝑥(𝑥 + 4) = 12
Sol: 𝑥(𝑥 + 4) = 12
𝑥² + 4𝑥 − 12 = 0
𝑥² − 2𝑥 + 6𝑥 − 12 = 0
𝑥(𝑥 − 2) + 6(𝑥 − 2) = 0
(𝑥 − 2)(𝑥 + 6) = 0
𝑥 − 2 = 0 𝑜𝑟 𝑥 + 6 = 0
𝑥 = 2 𝑜𝑟 𝑥 = −6
𝑇ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 𝑥(𝑥 + 4) = 12 𝑎𝑟𝑒 2 𝑎𝑛𝑑 − 6.
(vii) 3𝑥² − 5𝑥 + 2 = 0
Sol: 3𝑥² − 5𝑥 + 2 = 0
3𝑥² − 3𝑥 − 2𝑥 + 2 = 0
3𝑥(𝑥 − 1) − 2(𝑥 − 1) = 0
(𝑥 − 1)(3𝑥 − 2) = 0
𝑥 − 1 = 0 𝑜𝑟 3𝑥 − 2 = 0
𝑥 = 1 𝑜𝑟 𝑥 =2/3
𝑇ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 3𝑥²− 5𝑥 + 2 = 0 𝑎𝑟𝑒 1 𝑎𝑛𝑑 2/3
.

(ix) 3(𝑥 − 4)²− 5(𝑥 − 4) = 12
Sol: 3(𝑥 − 4)² − 5(𝑥 − 4) = 12
3(𝑥² − 8𝑥 + 16) − 5𝑥 + 20 − 12 = 0
3𝑥² − 24𝑥 + 48 − 5𝑥 + 8 = 0
3𝑥² − 29𝑥 + 56 = 0
3𝑥² − 21𝑥 − 8𝑥 + 56 = 0
3𝑥(𝑥 − 7) − 8(𝑥 − 7) = 0
(𝑥 − 7)(3𝑥 − 8) = 0
𝑥 − 7 = 0 𝑜𝑟 3𝑥 − 8 = 0
𝑥 = 7 𝑜𝑟 𝑥 =8/3
𝑇ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 3(𝑥 − 4)² − 5(𝑥 − 4) = 12 𝑎𝑟𝑒 7 𝑎𝑛𝑑 8/3

2.Find two numbers whose sum is 27 and product is 182.
Sol: Let one number = 𝑥, The second number= 27 − 𝑥
Product of numbers=182
𝑥(27 − 𝑥) = 182
27𝑥 − 𝑥² = 182
−𝑥² + 27𝑥 − 182 = 0
𝑥² − 27𝑥 + 182 = 0
𝑥² − 13𝑥 − 14𝑥 + 182 = 0
𝑥(𝑥 − 13) − 14(𝑥 − 13) = 0
(𝑥 − 13)(𝑥 − 14) = 0
𝑥 − 13 = 0 𝑜𝑟 𝑥 − 14 = 0
𝑥 = 13 𝑜𝑟 𝑥 = 14
If 𝑥 = 13 the required numbers are 13 and 14.
If 𝑥 = 14 the required numbers are 14 and 13.

3. Find two consecutive positive integers, sum of whose squares is 613.
Sol: Let the two consecutive positive integers be 𝑥, 𝑥 + 1.
Sum of whose squares = 613
𝑥² + (𝑥 + 1)² = 613
𝑥² + 𝑥² + 2𝑥 + 1 − 613 = 0
2𝑥² + 2𝑥 − 612 = 0
𝑥² + 𝑥 − 306 = 0
𝑥² − 17𝑥 + 18𝑥 − 306 = 0
𝑥(𝑥 − 17) + 18(𝑥 − 17) = 0
(𝑥 − 17)(𝑥 + 18) = 0
𝑥 = 17 𝑜𝑟 𝑥 = −18
∴ 𝑥 = 17 ( 𝑠𝑖𝑛𝑐𝑒 𝑥 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑠𝑜 𝑥 ≠ −18)
The required two consecutive positive integers are 17 , 18.

𝑥² + 𝑥² − 14𝑥 + 49 − 169 = 0
2𝑥² − 14𝑥 − 120 = 0
𝑥² − 7𝑥 − 60 = 0
𝑥² − 12𝑥 + 5𝑥 − 60 = 0
𝑥(𝑥 − 12) + 5(𝑥 − 12) = 0
(𝑥 − 12)(𝑥 + 5) = 0
𝑥 − 12 = 0 𝑜𝑟 𝑥 + 5 = 0
𝑥 = 12 𝑜𝑟 𝑥 = −5
∴ 𝑥 = 12 (𝑠𝑖𝑛𝑐𝑒 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑎 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑡𝑒𝑔𝑒𝑟 𝑠𝑜 𝑥 ≠ −5)

5.A cottage industry produces a certain number of pottery articles in a day. It was observed
on a particular day that the cost of production of each article (in rupees) was 3 more than
twice the number of articles produced on that day. If the total cost of production on that
day was Rs 90, find the number of articles produced and the cost of each article.
Sol: Let the number of articles produced=𝑥
The cost of each article=𝑅𝑠 (2𝑥 + 3)
Given the total cost of production on that day=Rs 90
𝑥(2𝑥 + 3) = 90
2𝑥² + 3𝑥 − 90 = 0
2𝑥² − 12𝑥 + 15𝑥 − 90 = 0
2𝑥(𝑥 − 6) + 15(𝑥 − 6) = 0
(𝑥 − 6)(2𝑥 + 15) = 0
𝑥 − 6 = 0 𝑜𝑟 2𝑥 + 15 = 0
𝑥 = 6 𝑜𝑟 𝑥 =−15/2
∴ 𝑥 = 6 ( 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑖𝑠 𝑎𝑙𝑤𝑎𝑦𝑠 𝑐𝑎𝑛′𝑡𝑏𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒)
The number of articles produced= 𝑥 =6
The cost of each article=𝑅𝑠 (2𝑥 + 3)=𝑅𝑠 (2 × 6 + 3) = 𝑅𝑠 15.

Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40
square meters.
Sol: Let the length of the rectangle=𝑙 , breadth =𝑏
Perimeter of the rectangle=28 m
2(𝑙 + 𝑏) = 28 ⇒ 𝑙 + 𝑏 =28/2
⇒ 𝑙 + 𝑏 = 14 ⇒ 𝑏 = 14 − 𝑙
Area of the square=40 square meters.
⇒ 𝑙 × 𝑏 = 40
⇒ 𝑙(14 − 𝑙) = 40
⇒ 14𝑙 − 𝑙² − 40 = 0
⇒ −𝑙² + 14𝑙 − 40 = 0
⇒ 𝑙² − 14𝑙 + 40 = 0
⇒ 𝑙² − 10𝑙 − 4𝑙 + 40 = 0
⇒ 𝑙(𝑙 − 10) − 4(𝑙 − 10) = 0
⇒ (𝑙 − 10)(𝑙 − 4) = 0
⇒ 𝑙 − 10 = 0 𝑜𝑟 𝑙 − 4 = 0
⇒ 𝑙 = 10 𝑜𝑟 𝑙 = 4
𝐼𝑓 𝑙 = 10 𝑚 𝑡ℎ𝑒𝑛 𝑏 = 14 − 10 = 4 𝑚
𝐼𝑓 𝑙 = 4 𝑚 𝑡ℎ𝑒𝑛 𝑏 = 14 − 4 = 10 𝑚
The dimensions of the rectangle are 10 m and 4 m.

The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm
then find its base and altitude.
Sol: Let altitude (h)=𝑥
The base of a triangle(b)=(𝑥 + 4)
The area of the triangle = 48 sq.cm
1/2× 𝑏 × ℎ = 48
(𝑥 + 4) × 𝑥 = 48 × 2
𝑥² + 4𝑥 − 96 = 0
𝑥² − 8𝑥 + 12𝑥 − 96 = 0
𝑥(𝑥 − 8) + 12(𝑥 − 8) = 0
(𝑥 − 8)(𝑥 + 12) = 0
𝑥 − 8 = 0 𝑜𝑟 𝑥 + 12 = 0
𝑥 = 8 𝑜𝑟 𝑥 = −12
∴ 𝑥 = 8 ( 𝐴𝑙𝑡𝑖𝑡𝑢𝑑𝑒 𝑐𝑎𝑛′𝑡𝑏𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒)
Altitude of the triangle=𝑥 = 8 𝑐𝑚
Base of the triangle=𝑥 + 4 = 8 + 4 = 12 𝑐𝑚.

8.Two trains leave a railway station at the same time. The first train travels towards west
and the second train towards north. The first train travels 5 km/hr faster than the second
train. If after two hours they are 50 km. apart find the average speed of each train.
Sol: Let the speed of second train=𝑥 𝑘𝑚/ℎ𝑟
The speed of first train=(𝑥 + 5)𝑘𝑚/ℎ𝑟
Time (t)=2 hr
Distance travelled by second train=𝑠 × 𝑡 = 2 × 𝑥 = 2𝑥 𝑘𝑚
Distance travelled by first train=𝑠 × 𝑡 = 2 × (𝑥 + 5) = (2𝑥 + 10)𝑘𝑚

In a class of 60 students, each boy contributed rupees equal to the number of girls and
each girl contributed rupees equal to the number of boys. If the total money then collected
was ₹1600. How many boys are there in the class?
Sol: Total number of students=60
Let the number of boys=𝑥
The number of girls=60 − 𝑥
Money contributed by the boys=𝑥 × (60 − 𝑥) = 60𝑥 − 𝑥²
Money contributed by the boys=(60 − 𝑥) × 𝑥 = 60𝑥 − 𝑥²
Total money collected=₹1600
60𝑥 − 𝑥² + 60𝑥 − 𝑥² = 1600
−2𝑥² + 120𝑥 − 1600 = 0
𝑥² − 60𝑥 + 800 = 0(𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑦 ′ − 2′)
𝑥² − 40𝑥 − 20𝑥 + 800 = 0
𝑥(𝑥 − 40) − 20(𝑥 − 40) = 0
(𝑥 − 40)(𝑥 − 20) = 0
𝑥 − 40 = 0 𝑜𝑟 𝑥 − 20 = 0
𝑥 = 40 𝑜𝑟 𝑥 = 20
∴ The number of boys in the class=40 or 20.

A motor boat heads upstream a distance of 24km on a river whose current is running at 3
km per hour. The trip up and back takes 6 hours. Assuming that the motor boat
maintained a constant speed, what was its speed?
Sol: Let the speed of the boat in water=𝑥 𝑘𝑚/ℎ𝑟
The speed of the river=3 𝑘𝑚/ℎ𝑟
The distance of river=24 km
The speed of the boat in upstream= (𝑥 − 3) 𝑘𝑚/ℎ𝑟

𝑥² − 8𝑥 − 9 = 0
𝑥² − 9𝑥 + 1𝑥 − 9 = 0
𝑥(𝑥 − 9) + 1(𝑥 − 9) = 0
(𝑥 − 9)(𝑥 + 1) = 0
𝑥 − 9 = 0 𝑜𝑟 𝑥 + 1 = 0
𝑥 = 9 𝑜𝑟 𝑥 = −1
∴ 𝑥 = 9 ( 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑎𝑡 𝑖𝑠 𝑐𝑎𝑛′𝑡 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒)
The speed of the boat in water= 9 km/hr.

Exercise 5.4 Solutions – Class X Mathematics

1. Find the nature of the roots of the following quadratic equations. If real roots exist, find them.

(i) \( 2x^2 – 3x + 5 = 0 \)

(ii) \( 3x^2 – 4\sqrt{3}x + 4 = 0 \)

(iii) \( 2x^2 – 6x + 3 = 0 \)

2. Find the values of \( k \) for each of the following quadratic equations, so that they have two equal roots.

(i) \( 2x^2 + kx + 3 = 0 \)

(ii) \( kx(x – 2) + 6 = 0 \), \( k \neq 0 \)

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

4. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the above situation possible? If so, determine their present ages.

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth. Comment on your answer.

Exercise 5.3 Solutions – Class X Mathematics

1. Find the roots of the following quadratic equations, if they exist.

(i) \( 2x^2 + x – 4 = 0 \)

(ii) \( 4x^2 + 4\sqrt{3}x + 3 = 0 \)

(iii) \( 5x^2 – 7x – 6 = 0 \)

(iv) \( x^2 + 5 = -6x \)

2. Find the roots of the quadratic equations given in Q.1 by applying the quadratic formula.

3. Find the roots of the following equations:

(i) \( x – \frac{1}{x} = 3 \), \( x \neq 0 \)

(ii) \( \frac{1}{x + 4} – \frac{1}{x – 7} = \frac{11}{30} \), \( x \neq -4, 7 \)

4. The sum of the reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is \( \frac{1}{3} \). Find his present age.

5. In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.

6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

9. Two water taps together can fill a tank in \( 9 \frac{3}{8} \) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

11. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

12. An object is thrown upwards with an initial velocity of 17 m/sec from a building with 12 m height. It is at a height of \( S = 12 + 17t – 5t^2 \) from the ground after a flight of \( t \) seconds. Find the time taken by the object to touch the ground.

13. If a polygon of \( n \) sides has \( \frac{1}{2} n(n-3) \) diagonals. How many sides are there in a polygon with 65 diagonals? Is there a polygon with 50 diagonals?

Exercise 5.2 Solutions – Class X Mathematics

1. Find the roots of the following quadratic equations by factorisation:

(i) \( x^2 – 3x – 10 = 0 \)

(ii) \( 2x^2 + x – 6 = 0 \)

(iii) \( \sqrt{2} x^2 + 7x + 5\sqrt{2} = 0 \)

(iv) \( 2x^2 – x + \frac{1}{8} = 0 \)

(v) \( 100x^2 – 20x + 1 = 0 \)

(vi) \( x(x + 4) = 12 \)

(vii) \( 3x^2 – 5x + 2 = 0 \)

(viii) \( x – \frac{3}{x} = 2 \) (\( x \neq 0 \))

(ix) \( 3(x – 4)^2 – 5(x – 4) = 12 \)

2. Find two numbers whose sum is 27 and product is 182.

3. Find two consecutive positive integers, sum of whose squares is 613.

4. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

6. Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.

7. The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm then find its base and altitude.

8. Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km apart, find the average speed of each train.

9. In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was Rs 1600, how many boys were there in the class?

10. A motor boat heads upstream a distance of 24 km in a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed in still water?

Exercise 5.1 Solutions – Class X Mathematics

1. Check whether the following are quadratic equations:

(i) \( (x + 1)^2 = 2(x – 3) \)

(ii) \( x^2 – 2x = (-2)(3 – x) \)

(iii) \( (x – 2)(x + 1) = (x – 1)(x + 3) \)

(iv) \( (x – 3)(2x + 1) = x(x + 5) \)

(v) \( (2x – 1)(x – 3) = (x + 5)(x – 1) \)

(vi) \( x^2 + 3x + 1 = (x – 2)^2 \)

(vii) \( (x + 2)^3 = 2x(x^2 – 1) \)

(viii) \( x^3 – 4x^2 – x + 1 = (x – 2)^3 \)

2. Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m². The length of the plot is one meter more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

QUADRATIC EQUATIONS Key Points

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EXERCISE-5.1

SOLUTIONS OF A QUADRATIC EQUATIONS BY FACTORISATION

EXERCISE-5.2

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