How do you use the quadratic formula to solve $x^{2} = 5 - x$ $x^2=5-x$ ?

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How do you use the quadratic formula to solve $x^{2} = 5 - x$ $x^2=5-x$ ?

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Answer 1 · 2016-07-27T18:26:19

$x = 1.79, x = - 2.79$ $x=1.79 , x=-2.79$

Explanation:

First, we need to get all the terms of the equations on one side of the equation
I added x to the left side
$x + x^{2} = 5$ $x+x^2=5$
then I subtracted 5
$- 5 + x + x^{2} = 0$ $-5+x+x^2=0$
now I will rearrange the equation to make it look similar to $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
$x^{2} + x - 5 = 0$ $x^2+x-5=0$

So we want to use the quadratic formula to find what the roots of x are
the quadratic formula is
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 \cdot a}$ $x=(-b+-sqrt(b^2-4ac))/(2*a)$

from the equation we will identify what $a, b, and c$ $a, b, and c$ are
$a = 1$ $a=1$
$b = 1$ $b=1$
$c = - 5$ $c=-5$

now we plug all this in to the formula
$x = \frac{- (1) \pm \sqrt{{(1)}^{2} - 4 (1) (- 5)}}{2 \cdot (1)}$ $x=(-(1)+-sqrt((1)^2-4(1)(-5)))/(2*(1))$

next we simplify
$x = \frac{- (1) \pm \sqrt{1 + 20}}{2}$ $x=(-(1)+-sqrt(1+20))/(2)$

$x = \frac{- 1 \pm \sqrt{21}}{2}$ $x=(-1+-sqrt(21))/(2)$

$x = \frac{- 1 \pm 4.58}{2}$ $x=(-1+-4.58)/(2)$

there will be two answers for x because of the $\pm$ $+-$
$x = \frac{- 1 + 4.58}{2} = \frac{3.58}{2} = 1.79$ $x=(-1+4.58)/(2)=3.58/2=1.79$
$x = \frac{- 1 - 4.58}{2} = - \frac{5.58}{2} = - 2.79$ $x=(-1-4.58)/(2)=-5.58/2=-2.79$

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How do you use the quadratic formula to solve $x^{2} = 5 - x$ $x^2=5-x$ ?

1 Answer

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How do you use the quadratic formula to solve x2=5−xx^2=5-x?

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How do you use the quadratic formula to solve $x^{2} = 5 - x$ $x^2=5-x$ ?