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

Answer: (A)

$4^{\left(x^2+2\right)}-9\cdot 2^{\left(x^2+2\right)}+8=0$
$\Rightarrow {\left(2^{\left(x^2+2\right)}\right)}^2-9\cdot 2^{\left(x^2+2\right)}+8=0$
Put $2^{\left(x^2+2\right)}=y$, 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.