Exercise 4.3: Quadratic Equations — NCERT Solutions

Here $a = 2$, $b = -6$, $c = 5$.

Discriminant: $D = b^2 - 4ac = (-6)^2 - 4(2)(5) = 36 - 40 = -4$.

Since $D < 0$, the equation has no real roots (the roots are imaginary/complex).

For two equal real roots, the discriminant must be zero: $b^2 - 4ac = 0$.

Here $a = k$, $b = -2\sqrt{5}$, $c = 4$.

$(-2\sqrt{5})^2 - 4(k)(4) = 0$ $4 \times 5 - 16k = 0$ $20 - 16k = 0$ $k = \frac{20}{16} = \frac{5}{4}$

Hence $k = \dfrac{5}{4}$.

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

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

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

The repeated root is: $x = \frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$

So both roots equal $\dfrac{2}{\sqrt{3}}$.

For real and equal roots, $b^2 - 4ac = 0$.

Here $a = 1$, $b = k$, $c = 9$.

$k^2 - 4(1)(9) = 0$ $k^2 - 36 = 0$ $k^2 = 36$ $k = \pm 6$

Hence $k = 6$ or $k = -6$.

Here $a = k+1$, $b = -2(k-1)$, $c = 1$.

For equal roots, $b^2 - 4ac = 0$: $[-2(k-1)]^2 - 4(k+1)(1) = 0$ $4(k-1)^2 - 4(k+1) = 0$

Divide by $4$: $(k-1)^2 - (k+1) = 0$ $k^2 - 2k + 1 - k - 1 = 0$ $k^2 - 3k = 0$ $k(k - 3) = 0$ $k = 0 \quad \text{or} \quad k = 3$

(Note: $k = -1$ would make it non-quadratic, so it is excluded, but here neither value is $-1$.) Hence $k = 0$ or $k = 3$.