Question

The equation $4^{(x^2+2)}-9\cdot 2^{(x^2+2)}+8=0$ has the solution

• A. $x=1$
• B. $x=0$
• C. $x=\sqrt{2}$
• D. $x=-\sqrt{2}$

Answer 1

The correct option is A $x=1$

$4^{(x^2+2)}-9\cdot 2^{(x^2+2)}+8=0$

$\Rightarrow {\left(2^{(x^2+2)}\right)}^2-9\cdot 2^{(x^2+2)}+8=0$

Put $2^{(x^2+2)}=y$, then $y^2-9y+8=0$ which gives $y=8$ and $y=1$.

When $y=8$, then $2^{x^2+2}=8$

$\Rightarrow 2^{x^2+2}=2^3$

$\Rightarrow x^2+2=3$

$\Rightarrow x^2=1$

$\Rightarrow x=1,-1$

When $y=1$, then $2^{x^2+2}=1$

$\Rightarrow 2^{x^2+2}=2^0$

$\Rightarrow x^2+2=0$

$\Rightarrow x^2=-2$

which is not possible.