How do you solve ${(3 x - 1)}^{2} = 5$ $(3x - 1) ^ { 2} = 5$ ?

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Answer 1 · 2017-03-17T02:29:02

$x = \frac{\pm \sqrt{5} + 1}{3}$ $x =(+- sqrt(5) +1)/ 3$

${(3 x - 1)}^{2} = 5$ $(3x -1)^2 = 5$

$3 x - 1 = \pm \sqrt{5}$ $3x -1 = +- sqrt(5)$

$3 x = \pm \sqrt{5} + 1$ $3x = +- sqrt(5) + 1$

$x = \frac{\pm \sqrt{5} + 1}{3}$ $x =(+- sqrt(5) + 1)/ 3$

How do you solve (3x - 1) ^ { 2} = 5(3x−1)2=5(3x - 1) ^ { 2} = 5?