How do you solve $y^{2} - 6 y - 4 = 0$ $y^2-6y-4=0$ by completing the square?

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How do you solve $y^{2} - 6 y - 4 = 0$ $y^2-6y-4=0$ by completing the square?

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Answer 1 · 2015-04-29T12:15:29

$y^{2} - 6 y - 4 = 0$ $y^2-6y-4 = 0$

Move the constant term to the right side:
$y^{2} - 6 y = 4$ $y^2-6y = 4$

Add the value of $a^{2}$ $a^2$ to both sides where
$2 a = - 6$ $2a = -6$
since we want ${(y + a)}^{2}$ $(y+a)^2$
$= y^{2} + 2 a x + a^{2}$ $= y^2 +2ax +a^2$
$= y^{2} - 6 y + a^{2}$ $= y^2-6y+a^2$

$a = - 3$ $a =-3$
$a^{2} = 9$ $a^2 = 9$

$y^{2} - 6 y + 9 = 4 + 9$ $y^2-6y+9= 4+9$

${(y - 3)}^{2} = 13$ $(y-3)^2 = 13$

$y - 3 = \pm \sqrt{13}$ $y-3 = +-sqrt(13)$

$y = 3 \pm \sqrt{13}$ $y = 3 +-sqrt(13)$

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How do you solve $y^{2} - 6 y - 4 = 0$ $y^2-6y-4=0$ by completing the square?

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How do you solve y2−6y−4=0y^2-6y-4=0 by completing the square?

1 Answer

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How do you solve $y^{2} - 6 y - 4 = 0$ $y^2-6y-4=0$ by completing the square?