# Solve this inequality \frac{1}{x^2-14x+40}\leq0

Solve this inequality
$\frac{1}{{x}^{2}-14x+40}\le 0$
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Giovanny Howe
Given:
$\frac{1}{{x}^{2}-14x+40}\le 0$
Factorising the term ${x}^{2}-14+40$
Let $p\left(x\right)={x}^{2}-14x+40$
$={x}^{2}-4x-10x+40$
$=x\left(x-4\right)-10\left(x-4\right)$
$=\left(x-10\right)\left(x-4\right)$
$p\left(x\right)=0$ when $x=\left(10,4\right)$
$p\left(x\right)>0$ when $x\in \left(-\mathrm{\infty },4\right)$
$p\left(x\right)<0$ when $x\in \left(4,10\right)$
$p\left(x\right)>0$ when $x\in \left(10,\mathrm{\infty }\right)$
For $\frac{1}{{x}^{2}-14x+40}\le 0$
${x}^{2}-14x+40$ should be less that 0. Thus, the solution is $\left(4,10\right)$