What is the sum of all real solutions to the equation \sqrt{2x + 15} = x? — GMAT

What is the sum of all real solutions to the equation $\sqrt{2x + 15} = x$ ?

The sum of all real solutions to the equation is 5.

Squaring both sides of

\sqrt{2x + 15} = x

yields

2x + 15 = x^2

, which rearranges to

x^2 - 2x - 15 = 0

. Factoring gives

(x - 5)(x + 3) = 0

, yielding candidate roots

x = 5

and

x = -3

. Substituting

x = 5

into the original equation gives

\sqrt{25} = 5

, which is true. Substituting

x = -3

gives

\sqrt{9} = -3

, which is false because the principal square root must be non-negative. Therefore,

x = 5

is the only valid real solution, and its sum is 5.

Square both sides of the equation to eliminate the radical.

2x + 15 = x^2

Squaring both sides removes the radical so the equation can be expressed in quadratic form.

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

x^2 - 2x - 15 = 0

Setting the quadratic expression equal to zero enables solution by factoring.

Factor the quadratic polynomial.

(x - 5)(x + 3) = 0 \implies x = 5 \text{ or } x = -3

Factoring determines the candidate roots of the quadratic equation.

Check candidate roots in the original equation

\sqrt{2x + 15} = x

for extraneous solutions.

For

x = 5

\sqrt{2(5) + 15} = \sqrt{25} = 5

(Valid). For

x = -3

\sqrt{2(-3) + 15} = \sqrt{9} = 3 \neq -3

(Extraneous).

Squaring both sides of an equation can introduce extraneous roots that violate the non-negativity constraint of principal square roots.

Sum all valid real solutions.

Sum = 5

Since

x = 5

is the only valid solution, the sum of all solutions is 5.

Quadratic Equations and Extraneous Solutions in Radical Equations

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