How do you solve $8 {(x + 4)}^{2} = 9$ $8(x+4)^2=9$ ?

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Answer 1 · 2016-07-23T14:11:57

$x = - 2.939 or x = - 5.061$ $x = -2.939 " or " x = -5.061$

Do not be tempted to multiply it all out and make equal to 0.

$8 {(x + 4)}^{2} = 9 divide by 8$ $8(x+4)^2 = 9" divide by 8"$

${(x + 4)}^{2} = \frac{9}{8} square root both sides$ $(x+4)^2 = 9/8" square root both sides"$ t8#

$x + 4 = \pm \sqrt{\frac{9}{8}}$ $x+4 = +-sqrt(9/8)$

$x = + \frac{3}{\sqrt{8}} - 4 or x = - \frac{3}{\sqrt{8}} - 4$ $x = +3/sqrt8 -4 " or "x = -3/sqrt8-4$

$x = - 2.939 or x = - 5.061$ $x = -2.939 " or " x = -5.061$

How do you solve 8(x+4)2=98(x+4)^2=9?