Question

Solve the radical equation. Which solution is extraneous? The solution x=-1 is an extraneous solution. Both x=-1 are x=-7 true solutions. The solution x=-7 is an extraneous solution. Neither x=-1 nor x=-7 is a true solution to the equation.

Answer 1

b 3dge

Explanation:

Both x=-1 are x=-7 true solutions.

Answer 2

Both x = -1 and x = -7 are true solutions.

Explanation:

Given

`x + 1 = √(-6x - 6)`

Required

Solve

`x + 1 = √(-6x - 6)`

Take square of both sides

`(x + 1)^2 = -6x - 6`

Open bracket

`x^2 + 2x + 1 = -6x -6`

Express as:

`x^2 + 2x+6x + 1 +6=0`

`x^2 + 8x + 7=0`

Expand

`x^2 + 7x +x + 7=0`

Factorize

`x(x + 7) + 1(x + 7) = 0`

Factor out x + 7

`(x + 1)(x + 7) = 0`

Solve:

`x + 1 =0\ or\ x + 7 = 0`

So:

`x = -1\ or\ x=-7`