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

Open in App

Solution

$x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$
Solving the Q.E in x,we have
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 6 \pm \sqrt{6^{2} - 4 (- (a^{2} + 2 a - 8))}}{2} = \frac{- 6 \pm \sqrt{36 + 4 a^{2} + 8 a - 32}}{2} = \frac{- 6 \pm \sqrt{4 a^{2} + 8 a + 4}}{2} = \frac{- 6 \pm \sqrt{4 (a + 1)^{2}}}{2}$
$x = \frac{- 6 \pm 2 (a + 1)}{2} = - 3 + a + 1 o r - 3 - a - 1 = a - 2 o r - a - 4$

Suggest Corrections

191

Similar questions

x^{2} + 6 x - (a^{2} + 2 a - 8) = 0

x^{2} + 6 x + 5 = 0

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

Solve each of the following quadratic equations:
$x^{2} + 12 x + 35 = 0$

Solve the following quadratic equation:
$x^{2} - 3 \sqrt{5} x + 10 = 0$

Join BYJU'S Learning Program