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

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

$$D = b^2 - 4ac$$

$$D = (-3)^2 - 4(2)(5) = 9 - 40 = -31$$$$\Rightarrow D = -31 < 0$$

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

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

$$D = b^2 - 4ac$$

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

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

The roots are:

$$x = \dfrac{-b}{2a}$$

$$x = \dfrac{4\sqrt{3}}{6}$$

$$\Rightarrow x = \dfrac{2}{\sqrt{3}}$$

$$\Rightarrow x = \dfrac{2\sqrt{3}}{3}$$

∴ The roots are $\dfrac{2}{\sqrt{3}}$ and $\dfrac{2}{\sqrt{3}}$.

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

$$D = b^2 - 4ac$$

$$D = (-6)^2 - 4(2)(3) = 36 - 24 = 12$$$$\Rightarrow D = 12 > 0$$

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

The roots are:

$$x = \dfrac{-b \pm \sqrt{D}}{2a}$$

$$x = \dfrac{6 \pm \sqrt{12}}{4}$$

$$\Rightarrow x = \dfrac{6 \pm 2\sqrt{3}}{4}$$

$$\Rightarrow x = \dfrac{3 \pm \sqrt{3}}{2}$$

∴ The roots are $\dfrac{3 + \sqrt{3}}{2}$ and $\dfrac{3 - \sqrt{3}}{2}$.

Here $a = 2,\ b = k,\ c = 3$.

For two equal roots, $D = 0$:

$$b^2 - 4ac = 0$$

$$k^2 - 4(2)(3) = 0$$

$$\Rightarrow k^2 - 24 = 0$$

$$\Rightarrow k^2 = 24$$

$$\Rightarrow k = \pm 2\sqrt{6}$$

∴ $k = \pm 2\sqrt{6}$

Writing the equation in the form $ax^2 + bx + c = 0$:

$$kx(x - 2) + 6 = 0$$

$$\Rightarrow kx^2 - 2kx + 6 = 0$$

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

For two equal roots, $D = 0$:

$$b^2 - 4ac = 0$$

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

$$\Rightarrow 4k^2 - 24k = 0$$

$$\Rightarrow 4k(k - 6) = 0$$

Setting each factor to zero:

$$4k = 0$$

$$\Rightarrow k = 0$$

$$k - 6 = 0$$

$$\Rightarrow k = 6$$

$k = 0$ is rejected, because the equation would no longer be quadratic.

∴ $k = \mathbf{6}$

Let the breadth be $x$ m.

The length is twice the breadth, so the length is $2x$ m.

The area is 800 m², so:

$$2x \times x = 800$$

$$\Rightarrow 2x^2 = 800$$

$$\Rightarrow 2x^2 - 800 = 0$$

Here $a = 2,\ b = 0,\ c = -800$.

$$D = b^2 - 4ac$$

$$D = 0^2 - 4(2)(-800) = 6400$$$$\Rightarrow D = 6400 > 0$$

∴ $D > 0$, so real roots exist and the design is possible.

Solving $2x^2 = 800$:

$$x^2 = 400$$

$$\Rightarrow x = 20 \quad \text{(taking the positive value)}$$

The length is $2x = 2(20) = 40$ m.

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

Let the age of one friend be $x$ years.

The sum of their ages is 20, so the other friend’s age is $(20 - x)$ years.

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

The product of their ages then was 48, so:

$$(x-4)(16-x) = 48$$

$$\Rightarrow 16x - x^2 - 64 + 4x = 48$$

$$\Rightarrow -x^2 + 20x - 64 = 48$$

$$\Rightarrow -x^2 + 20x - 112 = 0$$

$$\Rightarrow x^2 - 20x + 112 = 0$$

Here $a = 1,\ b = -20,\ c = 112$.

$$D = b^2 - 4ac$$

$$D = (-20)^2 - 4(1)(112) = 400 - 448 = -48$$$$\Rightarrow D = -48 < 0$$

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

The situation is not possible.

Let the length be $x$ m and the breadth be $y$ m.

The perimeter is 80 m, so:

$$2(x + y) = 80$$

$$\Rightarrow x + y = 40 \quad \cdots (1)$$

The area is 400 m², so:

$$xy = 400 \quad \cdots (2)$$

From equation (1):

$$x + y = 40$$

$$\Rightarrow x = 40 - y \quad \cdots (3)$$

Substituting $x = 40 - y$ in equation (2):

$$xy = 400$$

$$(40 - y)y = 400$$

$$\Rightarrow 40y - y^2 = 400$$

$$\Rightarrow y^2 - 40y + 400 = 0$$

Here $a = 1,\ b = -40,\ c = 400$.

$$D = b^2 - 4ac$$

$$D = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0$$$$\Rightarrow D = 0$$

∴ $D = 0$, so the equation has equal real roots and the design is possible.

The root is:

$$y = \dfrac{-b}{2a}$$

$$y = \dfrac{40}{2}$$

$$\Rightarrow y = 20$$

Substituting $y = 20$ in equation (3):

$$x = 40 - y$$

$$x = 40 - 20$$

$$\Rightarrow x = 20$$

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