Question

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

Answer 1

`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"`