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

Answer:

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

Step-by-step explanation:

Given

$x + 1 = \sqrt{-6x - 6}$

Required

Solve

$x + 1 = \sqrt{-6x - 6}$

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:

$x = -1\ or\ x=-7$