# Pre Calculus

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square root of 3x-4 = (x-4)^2-3

Normally, I either take the square root of both sides or I square both sides. Since there is a sqared and a square root, I am confused on how to go about it. Could you please head me in the right direction?

• Pre Calculus -

I don't see a square root anywhere.

Let's start by rewriting it as
3x -1 = (x-4)^2
Then turn it into a binomial equation with zero on one side.
3x -1 = x^2 -8x + 16
x^2 -11x + 17 = 0
That does not factor easily, so use the quadratic equation to solve.

x = (1/2)[11 +/- sqrt(121 - 68)]
= (1/2)[11 +/- sqrt53]
One of the roots is x = 9.14

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