If the roots of the equations $x^{2} + a^{2} = 8 x + 6 a$ are real, then find the interval of $a$ .

Open in App

Solution

$\begin{matrix} x^{2} + a^{2} = 8 x + 6 a x^{2} - 8 x + a^{2} - 6 a = 0 \end{matrix}$
Roots are reals,
Therefore,
$\begin{matrix} D \geq 0 64 - 4 (a^{2} - 6 a) \geq 0 16 - a^{2} + 6 a \geq 0 a^{2} - 6 a - 16 \leq 0 (a - 8) (a + 2) \leq 0 a \in (- 2, 8) \end{matrix}$

Suggest Corrections

Similar questions

x^{2} + a^{2} = 8 x + 6 a

a

x^{2} + a^{2} = 8 x + 6 a

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

a

(a^{2} - 4) x^{2} + (a^{2} - 6 a + 8) x + (a^{2} - 3 a + 2) = 0

a

Join BYJU'S Learning Program