What is the sum of all real solutions to the equation \sqrt{3x + 10} - x = 2? — GRE General Test

What is the sum of all real solutions to the equation $\sqrt{3x + 10} - x = 2$ ?

The sum of all real solutions is

2

To solve

\sqrt{3x + 10} - x = 2

, isolate the radical to get

\sqrt{3x + 10} = x + 2

. Squaring both sides yields

3x + 10 = x^2 + 4x + 4

. Rearranging into standard quadratic form gives

x^2 + x - 6 = 0

, which factors as

(x + 3)(x - 2) = 0

, yielding candidates

x = 2

and

x = -3

. Substituting

x = 2

into the original equation yields

\sqrt{16} - 2 = 2

, which is true. Substituting

x = -3

yields

\sqrt{1} - (-3) = 4 \neq 2

, so

x = -3

is extraneous. The only valid solution is

x = 2

, so the sum of all valid solutions is

2

Isolate the radical expression on one side of the equation

\sqrt{3x + 10} = x + 2

Isolating the radical allows squaring both sides cleanly to eliminate the radical sign.

Square both sides of the equation

3x + 10 = (x + 2)^2 = x^2 + 4x + 4

Squaring eliminates the square root, converting the equation into a polynomial form.

Rearrange terms into standard quadratic form

ax^2 + bx + c = 0

x^2 + x - 6 = 0

Grouping all terms on one side sets up the equation for factoring.

Factor the quadratic equation to find candidate solutions

(x + 3)(x - 2) = 0 \implies x = -3 \text{ or } x = 2

Factoring determines the values of

x

that solve the algebraic polynomial.

Test candidate solutions in the original radical equation to filter extraneous roots

For

x = 2

\sqrt{3(2) + 10} - 2 = \sqrt{16} - 2 = 4 - 2 = 2

(Valid). For

x = -3

\sqrt{3(-3) + 10} - (-3) = \sqrt{1} + 3 = 4 \neq 2

(Extraneous). Thus,

x = 2

is the only valid solution.

Squaring an equation can introduce extraneous roots that do not satisfy the original principal root definition.

Solving Radical Equations and Filtering Extraneous Roots

Estimated Time:1m 30s

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