How do you solve using the completing the square method $y^{2} + 16 y = 2$ $y^2 + 16y = 2$ ?

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How do you solve using the completing the square method $y^{2} + 16 y = 2$ $y^2 + 16y = 2$ ?

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Answer 1 · 2016-05-25T21:56:59

$y = - 8 \pm \sqrt{66}$ $y = -8 +- sqrt66$

Explanation:

find the number needed to complete the square:
${(\frac{b}{2})}^{2} = {(\frac{16}{2})}^{2} = 64$ $(b/2)^2 = (16/2)^2 = 64$
add to both sides of w=equation
$y^{2} + 16 y + 64 = 2 + 64$ $y^2 + 16y + 64 = 2 + 64$
factor and simplify
$(y + 8) (y + 8) = 66 \to {(y + 8)}^{2} = 66$ $(y + 8)(y + 8) = 66 -> (y + 8)^2 = 66$
take square root of both sides
$y + 8 = \pm \sqrt{66}$ $y + 8 = +- sqrt66$
subtract 8 from both sides
$y = - 8 \pm \sqrt{66}$ $y = -8 +-sqrt66$

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How do you solve using the completing the square method $y^{2} + 16 y = 2$ $y^2 + 16y = 2$ ?

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