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

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Answer 1 · 2015-05-23T22:28:46

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

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

Add $9$ to both ends to get:

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

So

$3x+1 = +-sqrt(9) = +-3$

Subtract $1$ from both sides to get:

$3x = -1+-3$

Divide both sides by $3$ to get:

$x=-1/3+-1$

That is $x = -4/3$ or $x = 2/3$

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

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