Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} - 4 a x - b^{2} + 4 a^{2} = 0$ [CBSE 2012]

Open in App

Solution

The given equation is $x^{2} - 4 a x - b^{2} + 4 a^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = −4a and C = $- b^{2} + 4 a^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 4 a)}^{2} - 4 \times 1 \times (- b^{2} + 4 a^{2}) = 16 a^{2} + 4 b^{2} - 16 a^{2} = 4 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 b^{2}} = 2 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 4 a) + 2 b}{2 \times 1} = \frac{4 a + 2 b}{2} = 2 a + b β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 4 a) - 2 b}{2 \times 1} = \frac{4 a - 2 b}{2} = 2 a - b$

Hence, $(2 a + b)$ and $(2 a - b)$ are the roots of the given equation.

Suggest Corrections

Similar questions

x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0

x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0

x^{2} + 5 x - (a^{2} + a - 6) = 0

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
$x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$

x^{2} + x + 2 = 0

Join BYJU'S Learning Program