predorsomwy
2022-04-26
Answered

Solve this inequality

$\frac{1}{{x}^{2}-14x+40}\le 0$

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Giovanny Howe

Answered 2022-04-27
Author has **18** answers

Given:

$\frac{1}{{x}^{2}-14x+40}\le 0$

Factorising the term${x}^{2}-14+40$

Let$p\left(x\right)={x}^{2}-14x+40$

$={x}^{2}-4x-10x+40$

$=x(x-4)-10(x-4)$

$=(x-10)(x-4)$

$p\left(x\right)=0$ when $x=(10,4)$

$p\left(x\right)>0$ when $x\in (-\mathrm{\infty},4)$

$p\left(x\right)<0$ when $x\in (4,10)$

$p\left(x\right)>0$ when $x\in (10,\mathrm{\infty})$

For$\frac{1}{{x}^{2}-14x+40}\le 0$

${x}^{2}-14x+40$ should be less that 0. Thus, the solution is $(4,10)$

Factorising the term

Let

For

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