How do you solve $x^{2} + 8 x = 20$ $x^2+8x=20$ using completing the square?

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Answer 1 · 2018-06-19T18:40:24

${(x + 4)}^{2} - 16 = 20$ $(x+4)^2-16=20$

${(x + 4)}^{2} = 36$ $(x+4)^2=36$

Square root both sides

$x + 4 = \pm 6$ $x+4=\pm6$

$x + 4 = 6 or x + 4 = - 6$ $x+4=6 or x+4=-6$

$x = 2 or x = - 10$ $x=2 or x=-10$

Answer 2 · 2018-06-19T18:50:58

$x = 2$ $x=2$ and $x = - 10$ $x=-10$

When we complete the square, we want to take half of our $b$ $b$ value, square it, and add it to the left side. Doing this, we get

$x^{2} + 8 x + 16 = 20 + 16$ $x^2+8x+color(blue)(16)=20+color(blue)(16)$

$16$ $16$ is the value ${(\frac{8}{2})}^{2}$ $(8/2)^2$ . Notice, we add it to both sides to maintain the equality.

Our equation can be further simplified as

${(x + 4)}^{2} = 36$ $(x+4)^2=36$

Taking the square root of both sides, we get

$x + 4 = 6$ $x+4=6$ and $x + 4 = - 6$ $x+4=-6$

Solving these equations gives us

$x = 2$ $x=2$ and $x = - 10$ $x=-10$

Hope this helps!

How do you solve x^2+8x=20x2+8x=20x^2+8x=20 using completing the square?