Question

The set of all $x$ satisfying $4^{x^2+2}-9\times 2^{x^2+2}+8=0$ consists of

• A. Infinitely many points,
• B. Finitely many points from the set of all natural numbers,
• C. Finitely many points from the set of all integers
• D. Two integers

Answer 1

Answer: (C), (D)

Two integers

$4^{x^2+2}-9\times 2^{x^2+2}+8=0$

Setting $2^{x^2+2}=t,$ the original equation reduces to $t^2-9t+8=0\Rightarrow t=1,8$.

If t =1 then $2^{x^2+2}=1\Rightarrow x^2+2=0$ but this has no solution.

If $2^{x^2+2}=8$ then $x^2+2=3\Rightarrow x=\pm 1.$