If x is the real solution to the equation \sqrt{5x + 6} - \sqrt{x + 3} = 3, what is the va… — SAT

If $x$ is the real solution to the equation $\sqrt{5x + 6} - \sqrt{x + 3} = 3$ , what is the value of $x$ ?

Answer: 6

The correct answer is

6

. Isolating the first radical gives

\sqrt{5x+6} = \sqrt{x+3} + 3

. Squaring both sides yields

5x+6 = x+12+6\sqrt{x+3}

, which simplifies to

4x-6 = 6\sqrt{x+3}

. Dividing by

2

and squaring both sides again gives

(2x-3)^2 = 9(x+3)

, leading to the quadratic equation

4x^2 - 21x - 18 = 0

. Factoring this equation yields

(4x+3)(x-6)=0

, which gives the potential solutions

x = 6

and

x = -\frac{3}{4}

. Checking both solutions in the original equation shows that

x = 6

is a valid solution because

\sqrt{36} - \sqrt{9} = 6 - 3 = 3

. The solution

x = -\frac{3}{4}

is extraneous because

\sqrt{2.25} - \sqrt{2.25} = 0 \neq 3

Isolate the first radical term and square both sides.

5x + 6 = x + 12 + 6\sqrt{x + 3}

Isolating one radical term simplifies the squaring process to eliminate that radical.

Isolate the remaining radical term and simplify the expression.

2x - 3 = 3\sqrt{x + 3}

Simplifying and isolating the remaining radical term prepares the equation for a second squaring step.

Square both sides to eliminate the remaining radical and form a quadratic equation.

4x^2 - 21x - 18 = 0

Squaring both sides removes the last radical, turning the radical equation into a polynomial equation.

Factor the quadratic equation to find all candidate solutions.

x = 6

x = -\frac{3}{4}

Factoring

(4x + 3)(x - 6) = 0

gives the roots of the quadratic equation.

Substitute candidate solutions back into the original equation to identify extraneous roots.

The only valid solution is

x = 6

; the solution

x = -\frac{3}{4}

is extraneous.

Squaring equations can introduce extraneous solutions, so candidate solutions must be verified in the original equation.

Solving equations with multiple radicals by isolating terms, squaring both sides, and verifying candidate solutions for extraneous roots.

Question