Selena Cowan

2022-01-21

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

KickAntitte06

Beginner2022-01-22Added 11 answers

The given expression is

$x}^{4$ + 16 = 0

Adding and subtracting$8{x}^{2}$ , we get

${x}^{4}-8{x}^{2}+8{x}^{2}+16=0$

${({x}^{2}+4)}^{2}-8{x}^{2}=0$

${({x}^{2}+4)}^{2}-{\left(2\sqrt{2x}\right)}^{2}=0$

$({x}^{2}+4+2\sqrt{2x})({x}^{2}+4-2\sqrt{2x})=0$

$(\because {a}^{2}-{b}^{2}=(a+b)(a-b))$

Solving the two quadratics separately.

$({x}^{2}+4+2\sqrt{2x})=0$

By quadratic formula

$x=\frac{-2\sqrt{2\pm \sqrt{8-4\times 4}}}{2}$

$=\frac{-2\sqrt{2\pm \sqrt{-8}}}{2}$

$=\frac{-2\sqrt{2\pm 2\sqrt{2i}}}{2}$

$=-\sqrt{2}\pm \sqrt{2i}$

Now , solving another quadratic equation, we get

$({x}^{2}+4-2\sqrt{2x})=0$

By quadratic formula

$x=\frac{2\sqrt{2\pm \sqrt{8-4\times 4}}}{2}$

$=\frac{2\sqrt{2\pm \sqrt{-8}}}{2}$

$=\frac{2\sqrt{2\pm 2\sqrt{2i}}}{2}$

$=\sqrt{2}\pm \sqrt{2i}$

Hence the roots of the given expression in rectangular form are

$-\sqrt{2}+\sqrt{2i}$

$-\sqrt{2}-\sqrt{2i}$

$\sqrt{2}+\sqrt{2i}$

$\sqrt{2}-\sqrt{2i}$

Adding and subtracting

Solving the two quadratics separately.

By quadratic formula

Now , solving another quadratic equation, we get

By quadratic formula

Hence the roots of the given expression in rectangular form are

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