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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.

Respuesta :

MrRoyal

Answer:

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

Step-by-step explanation:

Given

[tex]x + 1 = \sqrt{-6x - 6}[/tex]

Required

Solve

[tex]x + 1 = \sqrt{-6x - 6}[/tex]

Take square of both sides

[tex](x + 1)^2 = -6x - 6[/tex]

Open bracket

[tex]x^2 + 2x + 1 = -6x -6[/tex]

Express as:

[tex]x^2 + 2x+6x + 1 +6=0[/tex]

[tex]x^2 + 8x + 7=0[/tex]

Expand

[tex]x^2 + 7x +x + 7=0[/tex]

Factorize

[tex]x(x + 7) + 1(x + 7) = 0[/tex]

Factor out x + 7

[tex](x + 1)(x + 7) = 0[/tex]

Solve:

[tex]x + 1 =0\ or\ x + 7 = 0[/tex]

So:

[tex]x = -1\ or\ x=-7[/tex]

rshannon7127

Answer:

b   3dge

Step-by-step explanation:

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

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