How do you solve the equation ${(3 x - \frac{2}{3})}^{2} - 2 = \frac{5}{2}$ $(3x-2/3)^2-2=5/2$ ?

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How do you solve the equation ${(3 x - \frac{2}{3})}^{2} - 2 = \frac{5}{2}$ $(3x-2/3)^2-2=5/2$ ?

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Answer 1 · 2018-02-21T15:33:56

$x = \frac{2}{9} \pm \frac{\sqrt{2}}{2}$ $x=2/9+-sqrt2/2$

Explanation:

$add 2 to both sides$ $"add 2 to both sides"$

$\Rightarrow {(3 x - \frac{2}{3})}^{2} = \frac{9}{2}$ $rArr(3x-2/3)^2=9/2$

$take square root of both sides$ $color(blue)"take square root of both sides"$

$\sqrt{{(3 x - \frac{2}{3})}^{2}} = \pm \sqrt{\frac{9}{2}} \leftarrow note plus or minus$ $sqrt((3x-2/3)^2)=+-sqrt(9/2)larrcolor(blue)"note plus or minus"$

$\Rightarrow 3 x - \frac{2}{3} = \pm \frac{3}{\sqrt{2}} = \pm \frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3 \sqrt{2}}{2}$ $rArr3x-2/3=+-3/sqrt2=+-3/sqrt2xxsqrt2/sqrt2=(3sqrt2)/2$

$add \frac{2}{3} to both sides$ $"add "2/3" to both sides"$

$\Rightarrow 3 x = \frac{2}{3} \pm \frac{3 \sqrt{2}}{2}$ $rArr3x=2/3+-(3sqrt2)/2$

$divide both sides by 3$ $"divide both sides by 3"$

$\Rightarrow x = \frac{2}{9} \pm \frac{\sqrt{2}}{2} \leftarrow exact values$ $rArrx=2/9+-sqrt2/2larrcolor(red)"exact values"$

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How do you solve the equation ${(3 x - \frac{2}{3})}^{2} - 2 = \frac{5}{2}$ $(3x-2/3)^2-2=5/2$ ?

1 Answer

Explanation:

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How do you solve the equation ${(3 x - \frac{2}{3})}^{2} - 2 = \frac{5}{2}$ $(3x-2/3)^2-2=5/2$ ?