How do you solve ${(4 x + 3)}^{2} = 6$ $(4x + 3)^2 = 6$ ?

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Answer 1 · 2016-09-29T14:40:42

Take the square root then solve two linear equations.

Since we have a squared term, lets begin by taking the square root of both sides, so that we have:

$4 x + 3 = \pm \sqrt{6}$ $4x+3 = +-sqrt(6)$

Let's subtract 3 on both sides:

$4 x = \pm \sqrt{6} - 3$ $4x = +-sqrt(6) - 3$

Then divide by 4:

$x = \frac{\pm \sqrt{6} - 3}{4}$ $x = (+-sqrt(6)-3)/4$

So we have two solutions:

$x = \frac{\sqrt{6} - 3}{4}$ $x = (sqrt(6)-3)/4$

$x = \frac{- \sqrt{6} - 3}{4}$ $x = (-sqrt(6)-3)/4$

Including the $\pm$ $+-$ is important, or else we would have missed one of the solutions.

How do you solve (4x + 3)^2 = 6(4x+3)2=6(4x + 3)^2 = 6?