Selena Cowan

2022-01-21

Solve ${x}^{4}$ + 16 = 0 . Express the roots in rectangular form .

KickAntitte06

The given expression is
${x}^{4}$ + 16 = 0
Adding and subtracting $8{x}^{2}$ , we get
${x}^{4}-8{x}^{2}+8{x}^{2}+16=0$
${\left({x}^{2}+4\right)}^{2}-8{x}^{2}=0$
${\left({x}^{2}+4\right)}^{2}-{\left(2\sqrt{2x}\right)}^{2}=0$
$\left({x}^{2}+4+2\sqrt{2x}\right)\left({x}^{2}+4-2\sqrt{2x}\right)=0$
$\left(\because {a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)\right)$
$\left({x}^{2}+4+2\sqrt{2x}\right)=0$
$x=\frac{-2\sqrt{2±\sqrt{8-4×4}}}{2}$
$=\frac{-2\sqrt{2±\sqrt{-8}}}{2}$
$=\frac{-2\sqrt{2±2\sqrt{2i}}}{2}$
$=-\sqrt{2}±\sqrt{2i}$
Now , solving another quadratic equation, we get
$\left({x}^{2}+4-2\sqrt{2x}\right)=0$
$x=\frac{2\sqrt{2±\sqrt{8-4×4}}}{2}$
$=\frac{2\sqrt{2±\sqrt{-8}}}{2}$
$=\frac{2\sqrt{2±2\sqrt{2i}}}{2}$
$=\sqrt{2}±\sqrt{2i}$
Hence the roots of the given expression in rectangular form are
$-\sqrt{2}+\sqrt{2i}$
$-\sqrt{2}-\sqrt{2i}$
$\sqrt{2}+\sqrt{2i}$
$\sqrt{2}-\sqrt{2i}$

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