How do you complete the square for $y = - 3 x^{2} + 12 x + 9$ $y= -3x^2+12x+9$ ?

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Answer 1 · 2015-05-26T19:55:27

$y = - 3 x^{2} + 12 x + 9$ $y = -3x^2+12x+9$

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

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

$= - 3 ({(x - 2)}^{2} - 7)$ $= -3((x-2)^2 - 7)$

In general:
$a x^{2} + b x + c = a {(x + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $ax^2+bx+c = a(x+b/(2a))^2 + (c-b^2/(4a))$

Answer 2 · 2015-05-26T23:10:18

$y = - 3 x^{2} + 12 x + 9 = - 3 (x^{2} - 4 x - 3)$ $y = -3x^2 + 12x + 9 = -3(x^2 - 4x - 3)$ =

$= - 3 (x^{2} - 4 x + 4 - 4 - 3) = - 3 {(x - 2)}^{2} + 21$ $= -3(x^2 - 4x + 4 - 4 - 3) = -3(x - 2)^2 + 21$

${(x - 2)}^{2} = - \frac{21}{- 3} = 7$ $(x - 2)^2 = -21/-3 = 7$

$x - 2 = \sqrt{7} \to x = 2 + \sqrt{7}$ $x - 2 = sqrt7 -> x = 2 + sqrt7$

$x - 2 = - \sqrt{7} \to x = 2 - \sqrt{7}$ $x - 2 = -sqrt7 -> x = 2 - sqrt7$

How do you complete the square for y=−3x2+12x+9y= -3x^2+12x+9?