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

2 Answers
May 6, 2018

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

May 6, 2018

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:
Tony B