Posted by **Ash** on Sunday, September 6, 2009 at 12:40pm.

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 -
**drwls**, Sunday, September 6, 2009 at 2:18pm
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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