NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3#

Exercise 4.3 covers the nature of roots of a quadratic equation using the discriminant, and word problems that lead to quadratic equations.

Discriminant $D = b^2 - 4ac$ for $ax^2 + bx + c = 0$:

$D > 0$: Two distinct real roots
$D = 0$: Two equal real roots
$D < 0$: No real roots (roots are imaginary)

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

(i) $2x^2 – 3x + 5 = 0$

Here $a = 2,\ b = -3,\ c = 5$

$$D = b^2 - 4ac = 9 - 40 = -31$$

Since $D < 0$, the equation has no real roots.

(ii) $3x^2 – 4\sqrt{3}\,x + 4 = 0$

Here $a = 3,\ b = -4\sqrt{3},\ c = 4$

$$D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$$

Since $D = 0$, the equation has two equal real roots.

$$x = \frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$

∴ Roots are $\dfrac{2}{\sqrt{3}},\ \dfrac{2}{\sqrt{3}}$

(iii) $2x^2 – 6x + 3 = 0$

Here $a = 2,\ b = -6,\ c = 3$

$$D = 36 - 24 = 12 > 0$$

Since $D > 0$, the equation has two distinct real roots.

$$x = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$$

∴ Roots are $\dfrac{3 + \sqrt{3}}{2}$ and $\dfrac{3 - \sqrt{3}}{2}$

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, $D = 0$:

$$k^2 - 4(2)(3) = 0$$$$k^2 = 24$$$$k = \pm 2\sqrt{6}$$

(ii) $kx(x – 2) + 6 = 0$

Rewriting: $kx^2 - 2kx + 6 = 0$

Here $a = k,\ b = -2k,\ c = 6$

For equal roots, $D = 0$:

$$(-2k)^2 - 4(k)(6) = 0$$$$4k^2 - 24k = 0$$$$4k(k - 6) = 0$$

$k = 0$ (rejected, as equation becomes non-quadratic) or $k = 6$

∴ $k = \mathbf{6}$

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. #

Answer

Let breadth = $x$ m, then length = $2x$ m.

$$\text{Area} = 2x \times x = 2x^2 = 800$$$$x^2 = 400 \implies x = 20 \text{ m} \quad (\text{taking positive value})$$

Since $D = 0^2 - 4(2)(-800) = 6400 > 0$, real solution exists.

∴ Yes, it is possible. Length = 40 m and Breadth = 20 m.

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

Answer

Let the ages be $x$ and $(20 - x)$ years.

Four years ago their ages were $(x - 4)$ and $(16 - x)$.

$$(x-4)(16-x) = 48$$$$16x - x^2 - 64 + 4x = 48$$$$-x^2 + 20x - 64 = 48$$$$x^2 - 20x + 112 = 0$$$$D = 400 - 448 = -48 < 0$$

Since $D < 0$, the equation has no real roots.

∴ The situation is not possible.

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

Answer

Let length = $l$ m and breadth = $b$ m.

Perimeter: $2(l + b) = 80 \implies l + b = 40$

Area: $lb = 400$

From $l + b = 40$: $l = 40 - b$

$$(40 - b) \cdot b = 400$$$$40b - b^2 = 400$$$$b^2 - 40b + 400 = 0$$$$D = 1600 - 1600 = 0$$

Since $D = 0$, equal roots exist.

$$b = \frac{40}{2} = 20 \text{ m}, \quad l = 40 - 20 = 20 \text{ m}$$

∴ Yes, it is possible. Length = 20 m and Breadth = 20 m (square park).