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

2 Answers
May 6, 2018

Answer:

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

Answer:

#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