What is the set of all solutions to the equation \sqrt{x + 12} = x? — SAT

What is the set of all solutions to the equation $\sqrt{x + 12} = x$ ?

The solution set containing only

4

The correct answer is the set containing only

4

. Squaring both sides of the equation

\sqrt{x + 12} = x

produces

x + 12 = x^2

. Rearranging this into standard quadratic form gives

x^2 - x - 12 = 0

. Factoring the quadratic gives

(x - 4)(x + 3) = 0

, which yields potential solutions of

x = 4

and

x = -3

. Testing

x = 4

in the original equation yields

\sqrt{4 + 12} = \sqrt{16} = 4

, which is a true statement. Testing

x = -3

yields

\sqrt{-3 + 12} = \sqrt{9} = 3

, which does not equal

-3

. Thus,

x = -3

is extraneous, and

4

is the only valid solution.

Square both sides of the equation to eliminate the radical.

x + 12 = x^2

Squaring both sides of

\sqrt{x + 12} = x

removes the square root.

Rewrite the equation in standard quadratic form

ax^2 + bx + c = 0

x^2 - x - 12 = 0

Subtracting

x

and

12

from both sides moves all terms to one side of the equation.

Factor the quadratic equation.

(x - 4)(x + 3) = 0

We search for two numbers that multiply to

-12

and add to

-1

. These numbers are

-4

and

3

Solve for the potential solutions.

x = 4

x = -3

Setting each factor equal to zero yields the prospective values of

x

Check both potential solutions in the original equation to identify extraneous solutions.

Substituting

x = 4

\sqrt{4 + 12} = \sqrt{16} = 4

(valid). Substituting

x = -3

\sqrt{-3 + 12} = \sqrt{9} = 3 \neq -3

(extraneous).

Squaring both sides of an equation can introduce extraneous roots that must be discarded.

Solving radical equations and verifying solutions to eliminate extraneous roots

Question