How do you solve $x^{2} + 8 = 0$ $x^2+8=0$ using the quadratic formula?

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How do you solve $x^{2} + 8 = 0$ $x^2+8=0$ using the quadratic formula?

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Answer 1 · 2017-01-16T21:03:01

$x = 2 \sqrt{2} x i XXX or XXX x = - 2 \sqrt{2} x i$ $x=2sqrt(2)color(white)("x")icolor(white)("XXX")orcolor(white)("XXX")x=-2sqrt(2)color(white)("x")i$

(see below for use of the quadratic formula)

Explanation:

Standard quadratic form for an equation is:
$XXX a x^{2} + b x + c = 0$ $color(white)("XXX")color(red)ax^2+color(blue)bx+color(magenta)c=0$
and in this form the solutions are given by
$XXX x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(magenta)c))/(2color(red)a)$

Expressing the given equation into standard form:
$x^{2} + 8 = 0$ $x^2+8=0$
$XXX \to 1 x^{2} + 0 \cdot x + 8 = 0$ $color(white)("XXX")rarr color(red)1x^2+color(blue)0 * x +color(magenta)8 = 0$
with solutions:
$XXX x = \frac{- 0 \pm \sqrt{0^{2} - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$ $color(white)("XXX")x=(-color(blue)0+-sqrt(color(blue)0^2-4 * color(red)1 * color(magenta)8))/(2 * color(red)1)$

$XXX = \pm \frac{\sqrt{- 32}}{2} = \pm \frac{4 \sqrt{2} x i}{2} = \pm 2 \sqrt{2} x i$ $color(white)("XXX")=+-sqrt(-32)/2=+-(4sqrt(2)color(white)("x")i)/2=+-2sqrt(2)color(white)("x")i$

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How do you solve $x^{2} + 8 = 0$ $x^2+8=0$ using the quadratic formula?

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