Pre Calculus
posted by Ash .
square root of 3x4 = (x4)^23
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
I don't see a square root anywhere.
Let's start by rewriting it as
3x 1 = (x4)^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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