How do you solve $x^{2} + 8 = 0$ $x^2+8=0$ ?

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Answer 1 · 2018-03-20T08:21:34

Given: $x^{2} + 8 = 0$ $x^2+8=0$

Subtract $8$ $8$ from both sides.

$x^{2} = 0 - 8$ $x^2=0-8$

$= - 8$ $=-8$

Take square root of both sides.

$x = \sqrt{- 8}$ $x=sqrt(-8)$

$= \sqrt{4 \cdot - 2}$ $=sqrt(4*-2)$

$= \sqrt{4} \cdot \sqrt{- 2}$ $=sqrt(4)*sqrt(-2)$

$= 2 \sqrt{- 2}$ $=2sqrt(-2)$

$= 2 \sqrt{2} \cdot \sqrt{- 1}$ $=2sqrt(2)*sqrt(-1)$

$= 2 \sqrt{2} i$ $=2sqrt(2)i$

$= 2 i \sqrt{2}$ $=2isqrt(2)$

How do you solve x2+8=0x^2+8=0?