How do you solve $x^{2} + 5 x + 8 = 0$ $x^2+5x+8=0$ ?

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How do you solve $x^{2} + 5 x + 8 = 0$ $x^2+5x+8=0$ ?

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Answer 1 · 2016-04-30T12:04:13

$x = - \frac{5}{2} \pm \frac{\sqrt{7}}{2} i$ $x = -5/2+-sqrt(7)/2i$

Explanation:

Pre-multiply by $4$ $4$ (to avoid arithmetic with fractions), complete the square then use the difference of squares identity:

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

with $a = (2 x + 5)$ $a=(2x+5)$ and $b = \sqrt{7} i$ $b=sqrt(7)i$ as follows:

$0 = 4 (x^{2} + 5 x + 8)$ $0 = 4(x^2+5x+8)$

$= 4 x^{2} + 20 x + 32$ $=4x^2+20x+32$

$= {(2 x)}^{2} + 2 (2 x) (5) + 32$ $=(2x)^2+2(2x)(5)+32$

$= {(2 x + 5)}^{2} - 25 + 32$ $=(2x+5)^2-25+32$

$= {(2 x + 5)}^{2} + 7$ $=(2x+5)^2+7$

$= {(2 x + 5)}^{2} - {(\sqrt{7} i)}^{2}$ $=(2x+5)^2-(sqrt(7)i)^2$

$= ((2 x + 5) - \sqrt{7} i) ((2 x + 5) + \sqrt{7} i)$ $=((2x+5)-sqrt(7)i)((2x+5)+sqrt(7)i)$

$= (2 x + 5 - \sqrt{7} i) (2 x + 5 + \sqrt{7} i)$ $=(2x+5-sqrt(7)i)(2x+5+sqrt(7)i)$

So: $2 x = - 5 \pm \sqrt{7} i$ $2x = -5+-sqrt(7)i$

So: $x = - \frac{5}{2} \pm \frac{\sqrt{7}}{2} i$ $x = -5/2+-sqrt(7)/2i$

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Alternatively, use the quadratic formula:

The equation $x^{2} + 5 x + 8 = 0$ $x^2+5x+8 = 0$ is of the form $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$ with $a = 1$ $a=1$ , $b = 5$ $b=5$ and $c = 8$ $c=8$ .

This has roots given by the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$\frac{- 5 \pm \sqrt{5^{2} - (4 \cdot 1 \cdot 8)}}{2 \cdot 1}$ $(-5+-sqrt(5^2-(4*1*8)))/(2*1)$

$= \frac{- 5 \pm \sqrt{25 - 32}}{2}$ $=(-5+-sqrt(25-32))/2$

$= \frac{- 5 \pm \sqrt{- 7}}{2}$ $=(-5+-sqrt(-7))/2$

$= \frac{- 5 \pm \sqrt{7} i}{2}$ $=(-5+-sqrt(7)i)/2$

$= - \frac{5}{2} \pm \frac{\sqrt{7}}{2} i$ $=-5/2+-sqrt(7)/2i$

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