Question

What is `sqrt(4x-3)= 2+sqrt(2x-5)`?

Answer 1

`x={3,7}`

Explanation:

Given:

`sqrt(4x-3)=2+sqrt(2x-5)`

Square both sides:

`sqrt(4x-3)^2=(2+sqrt(2x-5))^2`

ACTUALLY square them:

`4x-3=4+4sqrt(2x-5)+2x-5`

Group like terms:

`2x-2=4sqrt(2x-5)`

Square both sides AGAIN:

`4x^2-8x+4=16(2x-5)`

Multiply:

`4x^2-8x+4=32x-80`

Group like terms:

`4x^2-40x+84=0`

Factor out `4`:

`4(x^2-10x+21)=0`

Then

`4(x^2 - 3x - 7x + 21) = 0`

`4[x(x-3)-7(x-3)] = 0`

So

`4(x-3)(x-7) = 0`

Answer 2

`x_1=3` and `x_2=7`

Explanation:

`sqrt(4x-3)=2+sqrt(2x-5)`

`sqrt(4x-3)-sqrt(2x-5)=2`

`(sqrt(4x-3)-sqrt(2x-5))^2=2^2`

`4x-3+2x-5-2sqrt(8x^2-26x+15)=4`

`6x-12=2sqrt(8x^2-26x+15)`

`3x-6=sqrt(8x^2-26x+15)`

`(3x-6)^2=8x^2-26x+15`

`9x^2-36x+36=8x^2-26x+15`

`x^2-10x+21=0`

`(x-3)*(x-7)=0`

Hence `x_1=3` and `x_2=7`