Question

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

Answer 1

x=4

Explanation:

to get rid of the square root on the right you square both sides so `2x+8 = x^2`
Then bring `2x+8` to the left so you have `0=(x^2)-2x+8`
now you can do the quadratic formula to find the zeroes

Answer 2

`x=4`

Explanation:

`color(blue)"square both sides"`

`"note that "sqrtaxxsqrta=(sqrta)^2=a`

`(sqrt(2x+8))^2=x^2`

`rArr2x+8=x^2`

`"rearrange into "color(blue)"standard form ";ax^2+bx+c=0`

`"subtract "2x+8" from both sides"`

`0=x^2-2x-8`

`"the factors of - 8 which sum to - 2 are - 4 and + 2"`

`rArr0=(x-4)(x+2)`

`"equate each factor to zero and solve for x"`

`x+2=0rArrx=-2`

`x-4=0rArrx=4`

`color(blue)"As a check"`

Substitute these values into the left side of the equation and if equal to the right side then they are the solution.

`x=-2tosqrt(-4+8)=sqrt4=2!=-2`

`rArrx=-2" is an extraneous solution"`

`x=4tosqrt(8+8)=sqrt16=4=" right side"`

`rArrx=4" is the solution"`

If you plot it as `y=0=x-sqrt(2x+8)` you get: