Question

How do you solve `sqrt (4x + 8) - 1 = x - 2`?

Answer 1

1) leave the square root alone on 1 side:
`sqrt(4x+8) = x-2+1 = x-1`

2) square both sides:
`4x+8 = (x-1)^2`

3) Expand the right side:`(a+b)^2=a^2+2ab+b^2`
`4x-8 = x^2 -2x+1`

4) Move everything to one side and you get:
`x^2 -6x +9 = 0`

5) This. luckily factors out to:
`(x-3)^2=0`
so `x=3`

Answer 2

`x=7,-1`

Explanation:

`sqrt(4x+8)-1=x-2`

`rarrsqrt(4x+8)=x-2+1`

`rarrsqrt(4x+8)=x-1`

Square both sides:

`rarr(sqrt(4x-8))^2=(x-1)^2`

Use the formula `(a+b)^2=a^2+2ab+b^2` for the equation on the R.h.s

`rarr4x-8=x^2-2x+1`

`rarr0=x^2-2x+1-4x8`

`rarr0=x^2-6x-7`

Now this is a quadratic equation:Solve using quadratic formula:

`rarrx=7,-1`