What is the only real solution to the equation \sqrt{3x + 13} = x + 3? — SAT

What is the only real solution to the equation $\sqrt{3x + 13} = x + 3$ ?

The correct answer is the value 1.

Squaring both sides of the equation

\sqrt{3x + 13} = x + 3

yields

3x + 13 = x^2 + 6x + 9

. Rearranging this into standard quadratic form gives

x^2 + 3x - 4 = 0

. Factoring the quadratic expression gives

(x + 4)(x - 1) = 0

, which yields the potential solutions

x = 1

and

x = -4

. Substituting

x = 1

back into the original equation gives

\sqrt{16} = 4

, which is true. Therefore, the value

1

is the only real solution.

Square both sides of the equation to eliminate the radical.

3x + 13 = (x + 3)^2

Squaring both sides eliminates the radical, allowing us to solve for

x

Expand the right side and rewrite the equation in standard quadratic form.

x^2 + 3x - 4 = 0

Expanding

(x+3)^2

yields

x^2 + 6x + 9

. Subtracting

3x

and

13

from both sides sets the quadratic equation to zero.

Factor the quadratic equation.

(x + 4)(x - 1) = 0

We find two numbers that multiply to

-4

and add to

3

, which are

4

and

-1

Solve for the potential solutions.

x = -4

x = 1

Setting each factor to zero yields the potential solutions.

Verify each potential solution in the original equation.

x = 1

is valid, and

x = -4

is extraneous.

Plugging in

x = 1

gives

\sqrt{16} = 4

, which is true. Plugging in

x = -4

gives

\sqrt{1} = -1

, which is false since the principal square root is non-negative.

Solving radical equations and checking for extraneous solutions.

Estimated Time:45s

Question