Quadratic Equations – Exercise 4.3

Nature of Roots

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

(i) $2x^2 - 3x + 5 = 0$

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

Question 1
(i) $2x^2 - 3x + 5 = 0$
$D = b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31$.
Since $D < 0$, no real roots exist.

(ii) $3x^2 - 4\sqrt{3}x + 4 = 0$
$D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$.
Real and equal roots exist. $x = \frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3}$.
✅ Roots are $\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}$.

(iii) $2x^2 - 6x + 3 = 0$
$D = (-6)^2 - 4(2)(3) = 36 - 24 = 12 > 0$.
Real and distinct roots exist. $x = \frac{-(-6) \pm \sqrt{12}}{2(2)} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$.
✅ Roots are $\frac{3 + \sqrt{3}}{2}$ and $\frac{3 - \sqrt{3}}{2}$.

Question 2
(i) $2x^2 + kx + 3 = 0$. For equal roots, $D = 0$.
$k^2 - 4(2)(3) = 0 \Rightarrow k^2 - 24 = 0 \Rightarrow k^2 = 24$.
$k = \pm\sqrt{24} = \pm 2\sqrt{6}$.
✅ $k = \pm 2\sqrt{6}$.

(ii) $kx(x - 2) + 6 = 0 \Rightarrow kx^2 - 2kx + 6 = 0$.
For equal roots, $D = 0 \Rightarrow (-2k)^2 - 4(k)(6) = 0$.
$4k^2 - 24k = 0 \Rightarrow 4k(k - 6) = 0$.
$k = 0$ or $k = 6$.
If $k=0$, the equation becomes $6=0$, which is false (and not quadratic).
✅ $k = 6$.

Word Problems

Solve the following problems:

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800 \text{ m}^2$? If so, find its length and breadth.

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.

5. Is it possible to design a rectangular park of perimeter 80 m and area $400 \text{ m}^2$? If so, find its length and breadth.

3. Let breadth $= x$ m. Then length $= 2x$ m.
Area $= x(2x) = 2x^2$.
Given Area $= 800 \Rightarrow 2x^2 = 800 \Rightarrow x^2 = 400$.
$x = \pm 20$. Since breadth cannot be negative, $x = 20$.
Length $= 2(20) = 40$ m.
✅ Yes, possible. Length = 40 m, Breadth = 20 m.

4. Let age of one friend $= x$. Age of other $= 20 - x$.
Four years ago: $(x - 4)$ and $(20 - x - 4) = (16 - x)$.
Product: $(x - 4)(16 - x) = 48$.
$16x - x^2 - 64 + 4x = 48 \Rightarrow -x^2 + 20x - 112 = 0$.
$x^2 - 20x + 112 = 0$.
Discriminant $D = (-20)^2 - 4(1)(112) = 400 - 448 = -48$.
Since $D < 0$, no real roots exist.
✅ No, the situation is not possible.

5. Perimeter $= 2(l + b) = 80 \Rightarrow l + b = 40 \Rightarrow b = 40 - l$.
Area $= l \times b = 400 \Rightarrow l(40 - l) = 400$.
$40l - l^2 = 400 \Rightarrow l^2 - 40l + 400 = 0$.
$(l - 20)^2 = 0 \Rightarrow l = 20$.
Breadth $b = 40 - 20 = 20$.
✅ Yes, possible. It is a square of side 20 m.

Quadratic Equations – Exercise 4.3

Overview

Nature of Roots

Word Problems