How do you solve $y^{2} - 12 y = - 35$ $y^2-12y=-35$ by completing the square?

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How do you solve $y^{2} - 12 y = - 35$ $y^2-12y=-35$ by completing the square?

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Answer 1 · 2018-05-09T08:36:17

${(y - 6)}^{2} - 1 = 0$ $(y-6)^2-1=0$

Explanation:

$y^{2} - 12 y + 35 - 0$ $y^2-12y+35-0$

${(y - 6)}^{2} + a = 0$ $(y-6)^2+a=0$

$y^{2} - 12 y + 36 + a = y^{2} - 12 y + 35$ $y^2-12y+36+a=y^2-12y+35$

$a = - 1$ $a=-1$

${(y - 6)}^{2} - 1 = 0$ $(y-6)^2-1=0$

Answer 2 · 2018-05-09T09:30:24

$y = 5 or y = 7$ $y=5" or "y=7$

Explanation:

$to solve using completing the square$ $"to solve using "color(blue)"completing the square"$

$add {(\frac{1}{2} coefficient of the y-term)}^{2} to both sides$ $" add "(1/2"coefficient of the y-term")^2" to both sides"$

$\Rightarrow y^{2} + 2 (- 6) y + 36 = - 35 + 36$ $rArry^2+2(-6)ycolor(red)(+36)=-35color(red)(+36)$

$\Rightarrow {(y - 6)}^{2} = 1$ $rArr(y-6)^2=1$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\sqrt{{(y - 6)}^{2}} = \pm \sqrt{1} \leftarrow note plus or minus$ $sqrt((y-6)^2)=+-sqrt1larrcolor(blue)"note plus or minus"$

$\Rightarrow y - 6 = \pm 1$ $rArry-6=+-1$

$add 6 to both sides$ $"add 6 to both sides"$

$\Rightarrow y = 6 \pm 1$ $rArry=6+-1$

$\Rightarrow y = 6 - 1 = 5 or y = 6 + 1 = 7$ $rArry=6-1=5" or "y=6+1=7$

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How do you solve $y^{2} - 12 y = - 35$ $y^2-12y=-35$ by completing the square?

2 Answers

Explanation:

Explanation:

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How do you solve y^2-12y=-35y2−12y=−35y^2-12y=-35 by completing the square?

2 Answers

Explanation:

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How do you solve $y^{2} - 12 y = - 35$ $y^2-12y=-35$ by completing the square?