How do you solve #(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:

#4x+3 = +-sqrt(6)#

Let's subtract 3 on both sides:

#4x = +-sqrt(6) - 3#

Then divide by 4:

#x = (+-sqrt(6)-3)/4#

So we have two solutions:

#x = (sqrt(6)-3)/4#

#x = (-sqrt(6)-3)/4#

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