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

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

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Answer 1 · 2017-01-05T00:26:52

$x = - \frac{5}{2} \pm \frac{\sqrt{33}}{2}$ $x = -5/2+-sqrt(33)/2$

Explanation:

The difference of squares identity can be written:

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

We can solve the given quadratic by completing the square and using the difference of squares identity with $a = (2 x + 5)$ $a=(2x+5)$ and $b = \sqrt{33}$ $b=sqrt(33)$ .

I will multiply it by $4$ $4$ to start with, to cut down on the need for arithmetic involving fractions:

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

$0 = 4 x^{2} + 20 x - 8$ $color(white)(0) = 4x^2+20x-8$

$0 = {(2 x)}^{2} + 2 (2 x) (5) + 25 - 33$ $color(white)(0) = (2x)^2+2(2x)(5)+25-33$

$0 = {(2 x + 5)}^{2} - {(\sqrt{33})}^{2}$ $color(white)(0) = (2x+5)^2-(sqrt(33))^2$

$0 = ((2 x + 5) - \sqrt{33}) ((2 x + 5) + \sqrt{33})$ $color(white)(0) = ((2x+5)-sqrt(33))((2x+5)+sqrt(33))$

$0 = (2 x + 5 - \sqrt{33}) (2 x + 5 + \sqrt{33})$ $color(white)(0) = (2x+5-sqrt(33))(2x+5+sqrt(33))$

$0 = 4 (x + \frac{5}{2} - \frac{\sqrt{33}}{2}) (x + \frac{5}{2} + \frac{\sqrt{33}}{2})$ $color(white)(0) = 4(x+5/2-sqrt(33)/2)(x+5/2+sqrt(33)/2)$

Hence:

$x = - \frac{5}{2} \pm \frac{\sqrt{33}}{2}$ $x = -5/2+-sqrt(33)/2$

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

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