Table of Contents
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
