How do you solve $x^{2} + 2 x - 7 = 17$ $x^2+2x-7=17$ by completing the square?

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How do you solve $x^{2} + 2 x - 7 = 17$ $x^2+2x-7=17$ by completing the square?

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Answer 1 · 2015-05-15T14:14:33

Take $- 7$ $-7$ to the right:
$x^{2} + 2 x = 17 + 7$ $x^2+2x=17+7$
$x^{2} + 2 x = 24$ $x^2+2x=24$
add and subtract $1$ $1$ to the left:
$x^{2} + 2 x + 1 - 1 = 24$ $x^2+2x+1-1=24$
$(x^{2} + 2 x + 1) - 1 = 24$ $(x^2+2x+1)-1=24$
${(x + 1)}^{2} - 1 = 24$ $(x+1)^2-1=24$
${(x + 1)}^{2} = 25$ $(x+1)^2=25$ take the square root of both sides and get rid of the ${()}^{2}$ $()^2$ :
$x + 1 = \pm \sqrt{25}$ $x+1=+-sqrt(25)$
$x = - 1 \pm 5$ $x=-1+-5$
So you get:
$x_{1} = - 1 + 5 = 4$ $x_1=-1+5=4$
$x_{2} = - 1 - 5 = - 6$ $x_2=-1-5=-6$

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How do you solve $x^{2} + 2 x - 7 = 17$ $x^2+2x-7=17$ by completing the square?

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