Question

2^2x-3×2^(x+2)+32=0 solvw quadratic equation.

Answer 1

Answer:

x=3  ,  x=2

Step-by-step explanation:

$2^{2x}-3\times \:2^{\left(x+2\right)}+32=0 \\ \\\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c\\2^{x+2}=2^x\times \:2^2\\2^{2x}-3\times \:2^x\times \:2^2+32=0\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=\left(a^b\right)^c\\\\2^{2x}=\left(2^x\right)^2\\\left(2^x\right)^2-3\times \:2^x\times \:2^2+32=0\\\\\mathrm{Rewrite\:the\:equation\:with\:}2^x=u\\\left(u\right)^2-3u\times \:2^2+32=0\\\\\mathrm{Expand\:}u^2-3u\times \:2^2+32:\quad u^2-12u+32$[/tex]$u=\frac{-\left(-12\right)+\sqrt{\left(-12\right)^2-4\times \:1\times \:32}}{2\times \:1}:\quad 8\\u=\frac{-\left(-12\right)-\sqrt{\left(-12\right)^2-4\times \:1\times \:32}}{2\times \:1}:\quad 4\\\\\mathrm{The\:final\:solutions\:to\:the\:quadratic\:equation\:are:}\\u=8,\:u=4\\\mathrm{Since\:}u=2^x\mathrm{,\:solve\:the\:following\:in\:order\:to\:find\:}x\\\mathrm{Solve\:}\:2^x=8:\quad x=3\\\mathrm{Solve\:}\:2^x=4:\quad x=2\\x=3,x=2$

Answer 2

ANSWER:

You answer is in the attachment dear.