How do you solve using the completing the square method $y^{2} + 4 y - 45 = 0$ $y^2+4y-45=0$ ?

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How do you solve using the completing the square method $y^{2} + 4 y - 45 = 0$ $y^2+4y-45=0$ ?

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Answer 1 · 2016-11-23T16:06:49

$y = - 9 or y = 5$ $y=-9 or y=5$

Explanation:

To complete the square we use $\frac{1}{2}$ $1/2$ of the $y$ $y$ coefficient as follows:

$y^{2} + 4 y - 45 = 0$ $y^2 + 4y -45 = 0$
$:. (y^2 + 1/2(4))^2 - (1/2(4))^2 - 45 = 0$
$:. (y^2 + 2)^2 - (2)^2 - 45 = 0$
$:. (y^2 + 2)^2 - 4 - 45 = 0$
$:. (y^2 + 2)^2 - 49 = 0$
$:. (y^2 + 2)^2 = 49$

We can now take the square root remembering that when we do so we get two solutions as $a^2 = b <=> a = +- sqrtb$

$:. y^2 + 2 = +-7$
$:. y^2 = -2+-7$

Yielding the two solutions:

$y=-9 or y=5$

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How do you solve using the completing the square method $y^{2} + 4 y - 45 = 0$ $y^2+4y-45=0$ ?

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How do you solve using the completing the square method $y^{2} + 4 y - 45 = 0$ $y^2+4y-45=0$ ?