For all real values of x that satisfy the equation below, what is the value of x + 1? \sqr… — SAT

For all real values of

x

that satisfy the equation below, what is the value of

x + 1

\sqrt{5x + 39} - 3 = x

The correct answer is

6

By adding

3

to both sides, we get

\sqrt{5x + 39} = x + 3

. Squaring both sides yields

5x + 39 = x^2 + 6x + 9

. Setting this quadratic equation to zero gives

x^2 + x - 30 = 0

, which factors as

(x + 6)(x - 5) = 0

. The potential solutions are

x = 5

and

x = -6

. Checking these values in the original equation shows that only

x = 5

is valid:

\sqrt{5(5) + 39} - 3 = 8 - 3 = 5

, which satisfies the equation. The value

x = -6

is extraneous since

\sqrt{5(-6) + 39} - 3 = 3 - 3 = 0 \neq -6

. Therefore, the value of

x + 1

5 + 1 = 6

Isolate the radical expression on one side of the equation.

\sqrt{5x + 39} = x + 3

Isolating the radical allows us to square both sides to eliminate the square root.

Square both sides of the equation.

5x + 39 = (x + 3)^2 \implies 5x + 39 = x^2 + 6x + 9

Squaring a square root removes the radical, allowing us to solve the resulting quadratic equation.

Rearrange the terms to form a standard quadratic equation of the form

ax^2 + bx + c = 0

x^2 + x - 30 = 0

Subtracting

5x

and

39

from both sides sets the quadratic expression equal to zero.

Factor the quadratic equation to find potential solutions.

(x + 6)(x - 5) = 0 \implies x = -6 \text{ or } x = 5

Finding two numbers that multiply to

-30

and add up to

1

gives

6

and

-5

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

For

x = 5

\sqrt{5(5) + 39} - 3 = 8 - 3 = 5

(Valid). For

x = -6

\sqrt{5(-6) + 39} - 3 = 3 - 3 = 0 \neq -6

(Extraneous).

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

Use the valid solution to calculate the requested expression.

x + 1 = 5 + 1 = 6

The question asks for the value of

x + 1

Solving radical equations and identifying extraneous solutions

Question