How do you solve ${(- 2 x + 5)}^{2} = - 8$ $(-2x+5)^2 = -8$ ?

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How do you solve ${(- 2 x + 5)}^{2} = - 8$ $(-2x+5)^2 = -8$ ?

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Answer 1 · 2017-07-29T11:55:12

See a solution process below:

Explanation:

First, expand the term on the left using this special rule for multiplying quadratics:

${(x + y)}^{2} = x^{2} + 2 x y + y^{2}$ $(color(red)(x) + color(blue)(y))^2 = color(red)(x)^2 + 2color(red)(x)color(blue)(y) + color(blue)(y)^2$

Substituting gives:

${(- 2 x + 5)}^{2} = - 8$ $(color(red)(-2x) + color(blue)(5))^2 = -8$

${(- 2 x)}^{2} + (2 \cdot - 2 x \cdot 5) + 5^{2} = - 8$ $(color(red)(-2x))^2 + (2 * color(red)(-2x) * color(blue)(5)) + color(blue)(5)^2 = -8$

$4 x^{2} + (- 20 x) + 25 = - 8$ $4x^2 + (-20x) + 25 = -8$

$4 x^{2} - 20 x + 25 = - 8$ $4x^2 - 20x + 25 = -8$

We can next convert this to standard form:

$4 x^{2} - 20 x + 25 + 8 = - 8 + 8$ $4x^2 - 20x + 25 + color(red)(8) = -8 + color(red)(8)$

$4 x^{2} - 20 x + 33 = 0$ $4x^2 - 20x + 33 = 0$

We can now use the quadratic formula to find the solutions for $x$ $x$ . The quadratic formula states:

For $a x^{2} + b x + c = 0$ $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 \cdot a}$ $x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$4$ $color(red)(4)$ for $a$ $color(red)(a)$

$- 20$ $color(blue)(-20)$ for $b$ $color(blue)(b)$

$33$ $color(green)(33)$ for $c$ $color(green)(c)$ gives:

$x = \frac{- (- 20) \pm \sqrt{{- 20}^{2} - (4 \cdot 4 \cdot 33)}}{2 \cdot 4}$ $x = (-(color(blue)(-20)) +- sqrt(color(blue)(-20)^2 - (4 * color(red)(4) * color(green)(33))))/(2 * color(red)(4))$

$x = \frac{20 \pm \sqrt{400 - 528}}{8}$ $x = (color(blue)(20) +- sqrt(400 - 528))/8$

$x = \frac{20 \pm \sqrt{- 128}}{8}$ $x = (color(blue)(20) +- sqrt(-128))/8$

$x = \frac{20 \pm \sqrt{64 \cdot - 2}}{8}$ $x = (color(blue)(20) +- sqrt(64 * -2))/8$

$x = \frac{20 \pm \sqrt{64} \sqrt{- 2}}{8}$ $x = (color(blue)(20) +- sqrt(64)sqrt(-2))/8$

$x = \frac{20 \pm 8 \sqrt{- 2}}{8}$ $x = (color(blue)(20) +- 8sqrt(-2))/8$

Or

$x = \frac{20}{8} \pm \frac{8 \sqrt{- 2}}{8}$ $x = color(blue)(20)/8 +- (8sqrt(-2))/8$

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

Answer 2 · 2017-07-29T12:04:05

No Real solutions;
within Complex numbers: $x = \frac{2}{5} - \sqrt{2} i or \frac{2}{5} + \sqrt{2} i$ $x=2/5-sqrt(2)i" or "2/5+sqrt(2)i$

Explanation:

Given
$xxx {(- 2 x + 5)}^{2} = - 8$ $color(white)("xxx")(-2x+5)^2=-8$

We note that any Real value squared must be $\geq 0$ $>= 0$
therefore this equation has No valid Real solutions

If we are dealing with Complex values, then
$xxx {(- 2 x + 5)}^{2} = - 8$ $color(white)("xxx")(-2x+5)^2=-8$

$xxx 4 x^{2} - 20 x + 25 = - 8$ $color(white)("xxx")4x^2-20x+25=-8$

$xxx 4 x^{2} - 20 x + 33 = 0$ $color(white)("xxx")4x^2-20x+33=0$

Then applying the quadratic formula that tells us that an equation of the form: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
has solutions:
$xxx x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(white)("xxx")x=(-b+-sqrt(b^2-4ac))/(2a)$

We have solutions:
$xxx x = \frac{20 \pm \sqrt{400 - 4 \cdot 4 \cdot 33}}{2 \cdot 4}$ $color(white)("xxx")x=(20+-sqrt(400-4*4*33))/(2 * 4)$

$xxx = \frac{20 \pm \sqrt{- 128}}{8}$ $color(white)("xxx")=(20+-sqrt(-128))/8$

$xxx = \frac{20 \pm 8 \sqrt{2} i}{8}$ $color(white)("xxx")=(20+-8sqrt(2)i)/8$

$xxx = \frac{5}{2} \pm \sqrt{2} i$ $color(white)("xxx")=5/2+-sqrt(2)i$

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How do you solve ${(- 2 x + 5)}^{2} = - 8$ $(-2x+5)^2 = -8$ ?

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