Table of Contents
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
