How do you solve the quadratic equation by completing the square: $v^{2} + 6 v - 59 = 0$ $v^2+6v-59=0$ ?

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How do you solve the quadratic equation by completing the square: $v^{2} + 6 v - 59 = 0$ $v^2+6v-59=0$ ?

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Answer 1 · 2015-07-23T20:03:30

$^v^{2} + 6 v - 59 = 0$ $^v^2+6v-59 =0$
$XXXX$ $color(white)("XXXX")$ $\to$ $rarr$ $XXXX$ $color(white)("XXXX")$ $v = - 3 \pm 2 \sqrt{17}$ $v=-3+-2sqrt(17)$
$XXXX$ $color(white)("XXXX")$ (by completing the square)

Explanation:

$v^{2} + 6 v - 59 = 0$ $v^2+6v-59 = 0$
$XXXX$ $color(white)("XXXX")$ Move the constant to the right side (out of the way)
$v^{2} + 6 v = 59$ $v^2+6v = 59$
$XXXX$ $color(white)("XXXX")$ Add as the third term the value needed to make the right side a square
$v^{2} + 6 v + 3^{2} = 59 + 9$ $v^2+6v+3^2 = 59+9$
$XXXX$ $color(white)("XXXX")$ Re-write right side as a squared binomial
${(v + 3)}^{2} = 68$ $(v+3)^2 = 68$
$XXXX$ $color(white)("XXXX")$ Take the square root of both sides
$v + 3 = \pm \sqrt{68} = \pm 2 \sqrt{17}$ $v+3 = +-sqrt(68) = +-2sqrt(17)$
$XXXX$ $color(white)("XXXX")$ Isolate $v$ $v$ by subtracting $3$ $3$ from both sides
$v = - 3 \pm 2 \sqrt{17}$ $v = -3+-2sqrt(17)$

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How do you solve the quadratic equation by completing the square: $v^{2} + 6 v - 59 = 0$ $v^2+6v-59=0$ ?

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Explanation:

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How do you solve the quadratic equation by completing the square: v^2+6v-59=0v2+6v−59=0v^2+6v-59=0?

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Explanation:

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How do you solve the quadratic equation by completing the square: $v^{2} + 6 v - 59 = 0$ $v^2+6v-59=0$ ?