If x is a real number, what is the sum of all valid real solutions to the equation \sqrt{3… — GMAT

If $x$ is a real number, what is the sum of all valid real solutions to the equation $\sqrt{3x - 2} - \sqrt{x + 3} = 1$ ?

The sum of all valid real solutions is 6.

To solve the radical equation, we isolate

\sqrt{3x - 2}

to get

\sqrt{3x - 2} = 1 + \sqrt{x + 3}

. Squaring both sides yields

3x - 2 = x + 4 + 2\sqrt{x + 3}

, which simplifies to

x - 3 = \sqrt{x + 3}

. Squaring both sides a second time produces the quadratic

x^2 - 7x + 6 = 0

, giving candidate solutions

x = 6

and

x = 1

. Checking in the original equation reveals that

x = 6

yields

\sqrt{16} - \sqrt{9} = 1

, which is true, whereas

x = 1

yields

\sqrt{1} - \sqrt{4} = -1 \neq 1

, making

x = 1

extraneous. The only valid solution is 6, so the sum of all valid solutions is 6.

Isolate one radical expression on one side of the equation.

\sqrt{3x - 2} = 1 + \sqrt{x + 3}

Isolating a radical allows for clean squaring of both sides without creating cross-terms with multiple roots.

Square both sides of the equation.

3x - 2 = 1 + 2\sqrt{x + 3} + (x + 3) \implies 3x - 2 = x + 4 + 2\sqrt{x + 3}

Eliminates the radical on the left side.

Isolate the remaining radical term and simplify.

2x - 6 = 2\sqrt{x + 3} \implies x - 3 = \sqrt{x + 3}

Simplifies the algebraic expression prior to squaring a second time.

Square both sides again to obtain a quadratic equation.

(x - 3)^2 = x + 3 \implies x^2 - 6x + 9 = x + 3 \implies x^2 - 7x + 6 = 0

Eliminates all remaining radical terms.

Factor the quadratic equation to find algebraic candidate solutions.

(x - 6)(x - 1) = 0 \implies x = 6 \text{ or } x = 1

Identifies potential real roots.

Test candidate solutions in the original equation

\sqrt{3x - 2} - \sqrt{x + 3} = 1

to remove extraneous solutions.

For

x = 6

\sqrt{16} - \sqrt{9} = 4 - 3 = 1

(Valid). For

x = 1

\sqrt{1} - \sqrt{4} = 1 - 2 = -1 \neq 1

(Extraneous). Thus,

x = 6

is the sole valid solution.

Squaring equations during algebraic steps can introduce extraneous roots that do not satisfy the original radical equation.

Solving Radical Equations and Eliminating Extraneous Solutions

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