Question

If $\alpha ,\beta$ are the roots of the equation $4^x-3\left(2^{x+3}\right)+128=0$, then ${\alpha }^2+{\beta }^2=$

• A. $25$
• B. $16$
• C. $24$
• D. $7$

Answer 1

The correct option is A $25$

$4^x-3\left(2^{x+3}\right)+128=0\Rightarrow 2^{2x}-24\cdot 2^x+128=0$

Now put $2^x=y$
$\Rightarrow y^2-24y+128=0\Rightarrow y=16,8$

Now when
$2^x=16\Rightarrow x=4$

and when $2^x=8\Rightarrow x=3$

$\therefore {\alpha }^2+{\beta }^2=25$