How do you solve #sqrt(8x+1)=x+2#?

1 Answer
Mar 25, 2018

Answer:

#x=1" or "x=3#

Explanation:

#color(blue)"square both sides"#

#(sqrt(8x+1))^2=(x+2)^2#

#rArr8x+1=x^2+4x+4#

#"rearrange into "color(blue)"standard form"#

#rArrx^2-4x+3=0#

#"The factors of + 3 which sum to - 4 are - 1 and - 3"#

#rArr(x-1)(x-3)=0#

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

#x-1=0rArrx=1#

#x-3=0rArrx=3#

#color(blue)"As a check"#

Substitute these values into the equation and if both sides are equal then they are the solutions.

#x=1rArrsqrt9=3" and "1+2=3 color(white)(x)✔︎#

#x=3rArrsqrt25=5" and "3+2=5color(white)(x)✔︎#

#rArrx=1" or "x=3" are the solutions"#