Juan Hewlett

2021-12-29

Show the work steps for finding the roots of ${x}^{4}-13{x}^{2}+36$ algebraically.

sukljama2

We finding the roots of ${x}^{4}-13{x}^{2}+36$
Now ${x}^{4}-13{x}^{2}+36$
$={x}^{4}-9{x}^{2}-4{x}^{2}+36$
$={x}^{2}\left({x}^{2}-9\right)-4\left({x}^{2}-9\right)$
$=\left({x}^{2}-9\right)\left({x}^{2}-4\right)$
$\left[\therefore {a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$
$=\left({x}^{2}-{3}^{2}\right)\left({x}^{2}-{2}^{2}\right)$
$=\left(x+3\right)\left(x-3\right)\left(x+2\right)\left(x-2\right)$
For the roots of ${x}^{4}-13{x}^{2}+36$
${x}^{4}-13{x}^{2}+36=0$
$⇒\left(x+3\right)\left(x-3\right)\left(x+2\right)\left(x-2\right)=0$

$⇒x=-3$
$⇒x-3=0$
$⇒x=3$
$⇒x+2=0$
$x=-2$
$x-2=0$
$x=2$
Hence $x=-3,-2,2,3$

Janet Young

${x}^{4}-13{x}^{2}+36$
$\left(x-3\right)\left({x}^{3}+3{x}^{2}-4x-12\right)$
$\left(x-3\right)\left(x-2\right)\left({x}^{2}+5x+6\right)$
$\left(x-3\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)$

karton

Given:

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