What is the set of all real solutions to the equation below? \sqrt{3x+10} = x+2 — SAT

What is the set of all real solutions to the equation below?

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

The set containing only 2

The correct answer is the set containing only 2. Squaring both sides of the equation yields

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

. Rearranging the terms to set the equation to zero gives

x^2 + x - 6 = 0

. Factoring the quadratic expression gives

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

, indicating potential solutions of

x = -3

and

x = 2

. Substituting

x = 2

back into the original equation gives

\sqrt{16} = 4

, which is valid. Substituting

x = -3

back into the original equation gives

\sqrt{1} = -1

, which is invalid because the principal square root is non-negative. Thus,

x = -3

is extraneous and must be discarded.

Square both sides of the equation to eliminate the radical.

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

, which expands to

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

Squaring both sides is the standard algebraic method to isolate the term under a square root.

Move all terms to one side to form a quadratic equation equal to zero.

x^2 + x - 6 = 0

Setting the quadratic expression to zero allows us to solve it by factoring.

Factor the quadratic equation.

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

, which gives the potential solutions

x = -3

and

x = 2

Finding the factors helps us determine the roots of the equation.

Substitute both potential solutions back into the original equation to check for extraneous solutions.

For

x = 2

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

and

2+2 = 4

, which is true. For

x = -3

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

and

-3+2 = -1

, which is false.

Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.

Identifying extraneous solutions in radical equations

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