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

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Answer 1 · 2015-05-24T22:04:52

$0 = - \frac{2}{3} x^{2} - \frac{4}{3} x + 1 = - \frac{2}{3} (x^{2} + 2 x - \frac{3}{2})$ $0 = -2/3x^2-4/3x+1 = -2/3(x^2+2x-3/2)$

$= - \frac{2}{3} (x^{2} + 2 x + 1 - 1 - \frac{3}{2})$ $= -2/3(x^2+2x+1-1-3/2)$

$= - \frac{2}{3} ({(x + 1)}^{2} - \frac{5}{2})$ $= -2/3((x+1)^2-5/2)$

Divide both sides by $- \frac{2}{3}$ $-2/3$ to get:

$0 = {(x + 1)}^{2} - \frac{5}{2}$ $0 = (x+1)^2 - 5/2$

Add $\frac{5}{2}$ $5/2$ to both sides to get:

${(x + 1)}^{2} = \frac{5}{2}$ $(x+1)^2 = 5/2$

Then

$x + 1 = \pm \sqrt{\frac{5}{2}}$ $x+1 = +-sqrt(5/2)$

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

$x = - 1 \pm \sqrt{\frac{5}{2}} = - 1 \pm \frac{\sqrt{10}}{2}$ $x = -1 +- sqrt(5/2) = -1 +-sqrt(10)/2$

How do you solve (−23)x2+(−43)x+1=0(-2/3)x^2 + (-4/3)x + 1 = 0 by completing the square?