How do you complete the square to solve $0 = 9 x^{2} + 6 x - 8$ $0=9x^2 + 6x - 8$ ?

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Answer 1 · 2015-05-23T22:28:46

$0 = 9 x^{2} + 6 x - 8 = 9 x^{2} + 6 x + 1 - 1 - 8$ $0 = 9x^2+6x-8 = 9x^2+6x+1-1-8$

$= {(3 x + 1)}^{2} - 9$ $= (3x+1)^2-9$

Add $9$ $9$ to both ends to get:

${(3 x + 1)}^{2} = 9$ $(3x+1)^2 = 9$

So

$3 x + 1 = \pm \sqrt{9} = \pm 3$ $3x+1 = +-sqrt(9) = +-3$

Subtract $1$ $1$ from both sides to get:

$3 x = - 1 \pm 3$ $3x = -1+-3$

Divide both sides by $3$ $3$ to get:

$x = - \frac{1}{3} \pm 1$ $x=-1/3+-1$

That is $x = - \frac{4}{3}$ $x = -4/3$ or $x = \frac{2}{3}$ $x = 2/3$

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

How do you complete the square to solve 0=9x^2 + 6x - 80=9x2+6x−80=9x^2 + 6x - 8?