This was on a recent math test of mine - any help? (RADICAL EXPRESSION)?

Question

Word for word:
Find the real valued solution(s) of 3 - 2rad(4x - 3) + x = 0, Check for extraneous root.

By "rad", I mean the square root sign, its just that I can't type it in here. The terms in the parenthesis would be under the radical sign.

Any help would really be appreciated, as I now am redoing the test as homework to see what I got wrong and how to fix it.

Answer 1

3 - 2√(4x - 3) + x = 0
2√(4x - 3) = x + 3
4(4x - 3) = x^2 + 6x + 9
16x - 12 = x^2 + 6x + 9
x^2 - 10x + 21
(x - 3)(x - 7) = 0
x = 3, 7

3 - 2√(4*3 - 3) + 3 =? 0
6 - 2√9 =? 0
6 - 6 =? 0

3 - 2√(4*7 - 3) + 7 =? 0
10 - 2√25 =? 0
10 - 10 = 0

both roots are good.

Answer 2

Move everything that is not part of the rad to the other side

rad(4x-3)= (-3/2)-(1/3)x

square both sides

4x-3 = [(-3/2)-(1/3)x ] ^ 2

4x-3 = 9/4 -x+(1/9)x^2

0=21/4-5x+(1/9)x^2

-b +- rad( b^2 -4ac)
-------------------------
2a

I'm wrong, See the one above.

Answer 3

2sqrt(4x-3)=3+x
square both sides
4(4x-3)=9+6x+x^2
16x-12=9+6x+x^2
x^2-10x+21=0
(x-7)(x-3)=0
x=7,3