How do you solve by completing the square for $y = 3 x^{2} + 24 x + 8$ $y=3x^2+24x+8$ ?

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How do you solve by completing the square for $y = 3 x^{2} + 24 x + 8$ $y=3x^2+24x+8$ ?

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Answer 1 · 2015-05-28T04:58:04

$y = 3 (x^{2} + 8 x + 16) - 48 + 8$ $y = 3(x^2 + 8x + 16) - 48 + 8$

$y = 3 {(x + 4)}^{2} - 40$ $y = 3(x + 4)^2 - 40$

${(x + 4)}^{2} = \frac{40}{3} \to (x + 4) = \pm \frac{2 \sqrt{10}}{\sqrt{3}}$ $(x + 4) ^2 = 40/3 -> (x + 4) = +- (2sqrt10)/(sqrt3)$ = $\pm \frac{2 \sqrt{30}}{3}$ $+- (2sqrt30)/3$

$x = - \frac{24}{6} \pm \frac{2 \sqrt{30}}{18} = - 6 \pm \frac{\sqrt{30}}{9}$ $x = -24/6 +- (2sqrt30)/18 = -6 +- (sqrt30)/9$

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How do you solve by completing the square for $y = 3 x^{2} + 24 x + 8$ $y=3x^2+24x+8$ ?

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