Question

Solve the following inequalities.
$x^2{2}^{2x}+9(x+2)2^x+8x^2⩽(x+2)2^{2x}+9x^2{2}^x+8x+16$

Answer 1

$x^2\left(2^{2x}\right)+9(x+2)2^x+8x^2\le (x+2)2^{2x}+9x^{2x}2+8x+16$

segregating the terms as follows

$\Rightarrow x^2\left(2^{3x}\right)+9(x+2)2^x+8x^2-(x+2)2^{2x}-9x^2{2}^x-8x-16\le 0$

$\Rightarrow 2^{2x}[x^2-x-2]+2^x\left[9\right(x+2)-9x^2]+8x^2-8x-16\le 0.$

$\Rightarrow 2^{2x}[x^2-x-2]+2^x[9x+18-9x^2]+8x^2-8x-16\le 0$

$\Rightarrow 2^{2x}[x^2-x-2]+2^x(-9)[x^2-x-2]+8[x^2-x-2]\le 0$

$\Rightarrow (x^2-x-2)(2^{2x}-9(2^x)+8)\le 0$

$\Rightarrow x^2-x-2\le 0$

$\Rightarrow x^2-2x+x-2\le 0$

$\Rightarrow x(x-2)+1(x-2)\le 0$

$\Rightarrow (x-2)(x+1)\Rightarrow 0$

$\Rightarrow x\le 2\left(or\right)x\le -1$

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

put $2^x=y\Rightarrow 2^{2x}{y}^2$

$\Rightarrow y^2-9y+8\le 0$

$y^2-8y-y+8\le 0$

$\Rightarrow y(y-8)-1(y-8)\le 0$

$(y-8)(y-1)\le 0\Rightarrow y\le 8ory\le 1$

$y\le 8ory\le 1$

$2^x\le 2^3$ $2^x\le 1$

$\Rightarrow \downarrow$ $\downarrow$

$x\le 3$ Not possible

$\therefore x\le -1,x\le 2orx\le 3$