Question

Solve by factoring: x⁴-8x²+12=0​

Answer 1

After solving by factoring $x^4-8x^2+12=0$ we get $x=\sqrt{2}\,\,,x=-\sqrt{2}\,\,,x=\sqrt{6}\,\,,x=-\sqrt{6}$

Step-by-step explanation:

We need to solve by factoring: $x^4-8x^2+12=0$

Let x^2 = u then x^4 = u^2

Putting values:

$u^2-8u+12=0$

Factorizing:

$u^2-6u-2u+12=0$

$u(u-6)-2(u-6)=0$

$(u-2)(u-6)=0$

$(u-2)=0\,\,and\,\,(u-6)=0$

$u=2\,\,and\,\,u=6$

Putting value of u=x^2

$x^2=2\,\,and\,\,x^2=6$

Taking square root:

$x=\pm \sqrt{2}\,\,and\,\,x=\pm \sqrt{6}$

$x=\sqrt{2}\,\,,x=-\sqrt{2}\,\,,x=\sqrt{6}\,\,,x=-\sqrt{6}$

So, After solving by factoring $x^4-8x^2+12=0$ we get $x=\sqrt{2}\,\,,x=-\sqrt{2}\,\,,x=\sqrt{6}\,\,,x=-\sqrt{6}$

Keywords: Solving Quadratic Equation

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