Which of the following represents the complete set of real solutions to the equation \sqrt… — ACT

Which of the following represents the complete set of real solutions to the equation $\sqrt{4x + 28} - x = 4$ ?

The correct answer is

2

only.

The correct answer is

2

only. Isolating the radical in the original equation gives

\sqrt{4x+28} = x+4

. Squaring both sides yields

4x+28 = x^2+8x+16

. Rearranging into standard quadratic form gives

x^2+4x-12 = 0

. Factoring the quadratic yields

(x+6)(x-2) = 0

, giving the candidate solutions

x = -6

and

x = 2

. Testing

x = 2

in the original equation gives

\sqrt{36} - 2 = 4

, which is valid. Testing

x = -6

gives

\sqrt{4} - (-6) = 8 \neq 4

, meaning

-6

is extraneous.

Isolate the radical expression on one side of the equation.

\sqrt{4x + 28} = x + 4

Isolating the radical allows us to square both sides directly to eliminate the square root.

Square both sides of the equation.

4x + 28 = x^2 + 8x + 16

Squaring a square root eliminates the radical, and squaring the binomial

(x+4)

yields

x^2 + 8x + 16

Move all terms to one side of the equation to form a standard quadratic equation.

x^2 + 4x - 12 = 0

Subtracting

4x

and

28

from both sides of the equation simplifies it to the form

ax^2 + bx + c = 0

Factor the quadratic equation to find potential solutions.

(x + 6)(x - 2) = 0

, which gives potential solutions

x = -6

and

x = 2

Factoring is a standard method to solve quadratic equations of this form.

Substitute the potential solutions back into the original equation to check for extraneous solutions.

For

x = 2

\sqrt{4(2)+28}-2 = 6-2 = 4

, which is true. For

x = -6

\sqrt{4(-6)+28}-(-6) = 2+6 = 8 \neq 4

, which is false.

Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original radical equation.

Solving radical equations by isolating the radical, squaring both sides, and identifying extraneous solutions.

Question