If $\sqrt{16 x^{2} - 48 x + 36} = 2 x + 12$ , then the sum of integral values of $x$ satisfying the equation is

9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $8$
$\sqrt{16 x^{2} - 48 x + 36} = 2 x + 12$
$\Rightarrow \sqrt{4 (4 x^{2} - 12 x + 9)} = 2 (x + 6)$
$\Rightarrow 2 \sqrt{(2 x - 3)^{2}} = 2 (x + 6)$
$\Rightarrow | 2 x - 3 | = x + 6 (∵ \sqrt{a^{2}} = | a |)$

Case $1 :$ If $x \geq \frac{3}{2}$
then $2 x - 3 = x + 6$
$\Rightarrow x = 9$

Case $2 :$ If $x < \frac{3}{2}$
then $3 - 2 x = x + 6$
$\Rightarrow x = - 1$

Sum $= 9 + (- 1) = 8$

Suggest Corrections

Similar questions

\sqrt{16 x^{2} - 48 x + 36} = 2 x + 12

x

x

[- 4, 100]

| 2 x \sqrt{(2 x - 1)^{2} |} = 1

x

\frac{| x + 9 |}{| x - 5 |} = 8

x

2^{2 x} - 8 \times 2^{x} = - 12

48 x - 8 = k (6 x - 1)

Join BYJU'S Learning Program