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

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

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Answer 1 · 2017-06-06T05:32:15

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

Explanation:

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

As $(- \frac{2}{3}) x^{2}$ $(-2/3)x^2$ is not a complete square, let us multiply each term by $- \frac{3}{2}$ $-3/2$ , so that we get $x^{2}$ $x^2$ , which is a complete square. Then our equation becomes

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

or $x^{2} + 2 x - \frac{3}{2} = 0$ $x^2+2x-3/2=0$

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

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

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

i.e. $(x + 1 - \sqrt{\frac{5}{2}}) (x + 1 + \sqrt{\frac{5}{2}}) = 0$ $(x+1-sqrt(5/2))(x+1+sqrt(5/2))=0$

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

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

1 Answer

Explanation:

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