How do you solve ${(6 x + 2)}^{2} + 4 = 28$ $(6x+2)^2+4=28$ ?

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How do you solve ${(6 x + 2)}^{2} + 4 = 28$ $(6x+2)^2+4=28$ ?

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Answer 1 · 2017-07-23T14:41:38

$x = - \frac{1}{3} \pm \frac{1}{3} \sqrt{6}$ $x=-1/3+-1/3sqrt6$

Explanation:

$isolate {(6 x + 2)}^{2}$ $"isolate " (6x+2)^2$

$subtract 4 from both sides$ $"subtract 4 from both sides"$

$\Rightarrow {(6 x + 2)}^{2} = 24$ $rArr(6x+2)^2=24$

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

$\sqrt{{(6 x + 2)}^{2}} = \pm \sqrt{24} \leftarrow note plus or minus$ $sqrt((6x+2)^2)=+-sqrt24larr" note plus or minus"$

$\Rightarrow 6 x + 2 = \pm 2 \sqrt{6} \leftarrow simplifying radical$ $rArr6x+2=+-2sqrt6larr" simplifying radical"$

$subtract 2 from both sides$ $"subtract 2 from both sides"$

$\Rightarrow 6 x = \pm 2 \sqrt{6} - 2$ $rArr6x=+-2sqrt6-2$

$divide both sides by 6$ $"divide both sides by 6"$

$\Rightarrow x = \frac{- 2 \pm 2 \sqrt{6}}{6}$ $rArrx=(-2+-2sqrt6)/6$

$\Rightarrow x = - \frac{1}{3} \pm \frac{1}{3} \sqrt{6} \leftarrow exact values$ $color(white)(rArrx)=-1/3+-1/3sqrt6larrcolor(red)" exact values"$

$x \approx 0.483 to 3 dec. places$ $x~~ 0.483" to 3 dec. places"$

$or x \approx - 1.15 to 2 dec. places$ $"or " x~~-1.15" to 2 dec. places"$

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How do you solve ${(6 x + 2)}^{2} + 4 = 28$ $(6x+2)^2+4=28$ ?

1 Answer

Explanation:

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How do you solve (6x+2)^2+4=28(6x+2)2+4=28(6x+2)^2+4=28?

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How do you solve ${(6 x + 2)}^{2} + 4 = 28$ $(6x+2)^2+4=28$ ?