Question

Solve the inequation
$2^{{2x}^2-10x+3}+6^{x^2-5x+1}\ge 3^{{2x}^2-10x+3}$
find the number of integers satisfying the inequation.

Answer 1

The given in equation is equivalent to
${8.2}^{2(x^2-5x)}+{6.2}^{x^2-5x}.3^{x^2-5x}-3^{2(x^2-5x)}\ge 0$
Let $2^{x^2-5x}=f\left(x\right)and{3}^{x^2-5x}=g\left(x\right)$
Then $8.f^2\left(x\right)+6f\left(x\right).g\left(x\right)-27g^2\left(x\right)\ge 0$
dividing in each by $g^2\left(x\right)$
then $8{\left(\frac{f\left(x\right)}{g\left(x\right)}\right)}^2+6\left(\frac{f\left(x\right)}{g\left(x\right)}\right)-27\ge 0andlet\frac{f\left(x\right)}{g\left(x\right)}=t(t>0)then{8t}^2+6t-27\ge 0\Rightarrow (t-3/2)(t+9/4)\ge 0\Rightarrow t\ge 3/2andt\ge -9/4$
the second inequation has no root $(\because t>0)$
From the first inequation
$t\ge 3/2{\left(\frac{2}{3}\right)}^{x^2-5x}\ge {\left(\frac{2}{3}\right)}^{-1}(\because \frac{2}{3}<1)\Rightarrow x^2-5x\le -1\Rightarrow x^2-5x+1\le 0$
$\therefore \frac{5-\sqrt{21}}{2}\le x\le \frac{5+\sqrt{21}}{2}$
Hence$xϵ[\frac{5-\sqrt{21}}{2},\frac{5+\sqrt{21}}{2}]$