For a real number x, the equation \sqrt{3x + 10} = x + 2 is given. What is the value of th… — SAT

For a real number $x$ , the equation $\sqrt{3x + 10} = x + 2$ is given. What is the value of the expression $x - 1$ ?

The correct answer is the value 1, which is obtained by evaluating the expression for the only valid solution to the equation.

To solve the given radical equation, square both sides to obtain a quadratic equation, which simplifies to

x^2 + x - 6 = 0

. Factoring this equation yields potential solutions of

x = 2

and

x = -3

. Substituting

x = 2

back into the original equation produces a true statement, confirming it as a valid solution. Substituting

x = -3

results in an inequality, identifying it as an extraneous solution. Evaluating the expression for the valid solution yields

2 - 1 = 1

Eliminate the radical by squaring both sides of the equation.

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

, which expands to

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

Squaring both sides removes the square root, allowing the equation to be solved algebraically.

Move all terms to one side to set the quadratic equation to zero.

x^2 + x - 6 = 0

A quadratic equation must be in the form

ax^2 + bx + c = 0

to solve it by factoring.

Factor the quadratic equation.

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

, giving potential solutions of

x = -3

and

x = 2

Factoring allows us to find the roots that satisfy the quadratic relationship.

Check both potential solutions in the original equation to identify any extraneous solutions.

For

x = 2

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

(valid). For

x = -3

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

(invalid). Thus,

x = -3

is extraneous and

x = 2

is the only valid solution.

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

Evaluate the requested expression using the valid solution.

x - 1 = 2 - 1 = 1

The question asks for the value of the expression rather than the value of the variable itself.

Radical and Rational Equations

Question