If \sqrt{3x + 19} - x = 3, what is the value of x? — ACT

If $\sqrt{3x + 19} - x = 3$ , what is the value of $x$ ?

The value of

x

is 2.

To solve the equation, isolate the radical to obtain

\sqrt{3x + 19} = x + 3

. Squaring both sides yields

3x + 19 = x^2 + 6x + 9

. Rearranging terms results in the quadratic equation

x^2 + 3x - 10 = 0

, which factors into

(x + 5)(x - 2) = 0

. The potential solutions are

x = 2

and

x = -5

. Testing these in the original equation shows that

2

is a valid solution because

\sqrt{3(2) + 19} - 2 = 3

. Testing

-5

results in

\sqrt{3(-5) + 19} - (-5) = 7

, which does not equal

3

. Thus, the only real solution is

2

Isolate the radical expression on one side of the equation.

\sqrt{3x + 19} = x + 3

Isolating the radical allows us to eliminate it by squaring both sides.

Square both sides of the equation.

3x + 19 = x^2 + 6x + 9

Squaring a square root removes the radical. Remember to expand

(x+3)^2

correctly as

x^2 + 6x + 9

Rearrange the terms into standard quadratic form,

ax^2 + bx + c = 0

x^2 + 3x - 10 = 0

Moving all terms to one side sets up the quadratic equation for factoring.

Factor the quadratic expression.

(x + 5)(x - 2) = 0

Factoring allows us to find the potential values of

x

by setting each factor to zero.

Solve for the potential values of

x

x = -5

x = 2

Setting

x + 5 = 0

gives

x = -5

, and setting

x - 2 = 0

gives

x = 2

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

For

x = 2

\sqrt{3(2) + 19} - 2 = \sqrt{25} - 2 = 5 - 2 = 3

(valid). For

x = -5

\sqrt{3(-5) + 19} - (-5) = \sqrt{4} + 5 = 2 + 5 = 7 \neq 3

(extraneous).

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

Solving radical equations and checking for extraneous solutions.

Question