# How to solve 2x+root of 3x-2=3?

*print*Print*list*Cite

### 3 Answers

Supposing that the equation that you want to solve is `2x + sqrt(3x-2) = 3` , then you need to isolate square root to the left side such that:

`sqrt(3x-2) = 3 - 2x`

You need to raise to square both side to remove the square root such that:

`3x - 2 = (3 - 2x)^2`

You need to expand the binomial such that:

`3x -2 = 9 - 12x + 4x^2`

`4x^2 - 12x + 9 - 3x + 2 = 0`

`4x^2 - 15x + 11 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (15+-sqrt(225 - 176))/8`

`x_(1,2) = (15+-sqrt49)/8`

`x_(1,2) = (15+-7)/8 =gt x_1 = (15+7)/8 = 11/4`

`x_2 = (15-7)/8 = 1`

Notice that both solutions check the square root.

**Hence, evaluating soultions to equation yields `x_1 = 11/4` and `x_2 = 1` .**

4x2-15+11 can also by factoring.

» 4x2-4x-11x+11

»4x(x-1)-11(x-1)

» (4x-11) (x-1)

4x-11=0 and x-1=0

4x=11 and x=1

X=11/4 and 1

this is how you solve a quadratic equation:

2x+ (3x-2)(3x-2)=3

2x+ 9x^2-6x-6x+4=3

9x^2-10x+1=0

9x^2-9x-x+1=0

x(9x-1) +1(-9x+1)=0

(9x-1)(x-1)=0

x=1/9 or x=1

therefore the exact roots are 1/9 and 1.