Question

How do you solve `sqrt[7x+2]-2=7x` and find any extraneous solutions?

Answer 1

`x = -1/7` or `x = -2/7`

Explanation:

Move the constant term

`sqrt(7x+2) = 7x + 2`

Square both sides

`(sqrt(7x+2)) ^ 2 = (7x + 2) ^ 2`

`7x+2 = 49x^2 + 28x + 4`

Move everything over

`0 = 49x ^2 +21x+2`

Factor using the quadratic formula

`x_(1,2) = (-b ± sqrt(b ^2 - 4ac)) / (2a)`

So

`x_(1, 2) = (-21 ± sqrt(21 ^2 - 4(49)(2))) / (2(49))`

Solve to get

`x = -1 / 7" "` or `" "x = -2 / 7`

To check for extraneous roots, plug the solutions back into the original equation.

`sqrt(7(-1/7)+2) = 7(-1/7) + 2`

`1 = 1 ->` the solution `(-1/7)` is correct!

Repeat for the other solution

`sqrt(7(-2/7)+2) = 7(-2/7) + 2`

`0 = 0 ->` the solution `(-2/7)` is also correct!

Extraneous solutions appear when your solutions don't satisfy the original equation.