Solve each of the following quadratic equations:

$x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0$ [CBSE 2015]

Open in App

Solution

We write, $- 2 a x = (2 b - a) x - (2 b + a) x$ as $x^{2} \times [- (4 b^{2} - a^{2})] = - (4 b^{2} - a^{2}) x^{2} = (2 b - a) x \times [- (2 b + a) x]$
$∴ x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0 \Rightarrow x^{2} + (2 b - a) x - (2 b + a) x - (2 b - a) (2 b + a) = 0 \Rightarrow x [x + (2 b - a)] - (2 b + a) [x + (2 b - a)] = 0 \Rightarrow [x + (2 b - a)] [x - (2 b + a)] = 0$
$\Rightarrow x + (2 b - a) = 0 or x - (2 b + a) = 0 \Rightarrow x = - (2 b - a) or x = 2 b + a \Rightarrow x = a - 2 b or x = a + 2 b$
Hence, $a - 2 b$ and $a + 2 b$ are the roots of the given equation.

Suggest Corrections

Similar questions

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

x^{2} + 3 \sqrt{3} x - 30 = 0

Solve the following quadratic equation for $x$ :

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

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

Solve each of the following quadratic equations:
$x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$

Join BYJU'S Learning Program