Question

Eloise started to solve a radical equation in this way: Square root of negative 2x plus 1 − 3 = x Square root of negative 2x plus 1 − 3 + 3 = x + 3 Square root of negative 2x plus 1 = x + 3 Square root of negative 2x plus 1 − 1 = x + 3 − 1 Square root of negative 2 x = x + 2 (Square root of negative 2 x)2 = (x − 4)2 −2x = x2 − 8x + 16 −2x + 2x = x2 + 8x + 16 + 2x 0 = x2 + 10x + 16 0 = (x + 2)(x + 8) x + 2 = 0 x + 8 = 0 x + 2 − 2 = 0 − 2 x + 8 − 8 = 0 − 8 x = −2 x = −8 Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make?

Answer 1

We need to subtract 1 after squaring both sides

Explanation:

Step1:
`√(-2x+1) -3 = x \\`

adding 3 both sides

`√(-2x+1) =3 + x \\`

we need to square here first and then we need to subtract

and this is the error which Eloise make in solving it .

Answer 2

Please see attached image for the choices and proper formatting of the problem.

What she did:

`√(-2x + 1 ) - 1 = x + 3 - 1 √(-2x) = x + 2`

What she should have done:

`( √(-2x + 1) ^(2) = (x + 3)^2 -2x + 1 = (x + 3)^2 -2x + 1 - 1 = (x + 3)^2 - 1 -2x = (x +3)^2 - 1`

The error that Eloise made in solving the radical equation was that she subtracted 1 before squaring both sides.

To add, the equation where at least one variable expression is fixed inside a radical, usually a square root is called a radical equation.