How do you solve $(-2/3)x^2 + (-4/3)x + 1 = 0$ by completing the square?

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Answer 1 · 2015-05-24T22:04:52

$0 = -2/3x^2-4/3x+1 = -2/3(x^2+2x-3/2)$

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

$= -2/3((x+1)^2-5/2)$

Divide both sides by $-2/3$ to get:

$0 = (x+1)^2 - 5/2$

Add $5/2$ to both sides to get:

$(x+1)^2 = 5/2$

Then

$x+1 = +-sqrt(5/2)$

Subtract $1$ from both sides to get:

$x = -1 +- sqrt(5/2) = -1 +-sqrt(10)/2$

How do you solve (-2/3)x^2 + (-4/3)x + 1 = 0(-2/3)x^2 + (-4/3)x + 1 = 0 by completing the square?