What is the solution set of the equation $ \sqrt{2x + 6} + 4 = x + 3 $?

B. \{5\}. Choice B is correct. Subtracting 4 from both sides of $ \sqrt{2x + 6} + 4 = x + 3 $ isolates the radical expression on the left side of the equation as follows: $ \sqrt{2x + 6} = x - 1 $. Squaring both sides of $ \sqrt{2x + 6} = x - 1 $ yields $ 2x + 6 = x^2 - 2x + 1 $. This equation can be rewritten as a quadratic equation in standard form: $ x^2 - 4x - 5 = 0 $. One way to solve this quadratic equation is to factor the expression $ x^2 - 4x - 5 $ by identifying two numbers with a sum of $-4$ and a product of $-5$. These numbers are $-5$ and $1$. So the quadratic equation can be factored as $(x - 5)(x + 1) = 0$. It follows that $5$ and $-1$ are the solutions to the quadratic equation. However, the solutions must be verified by checking whether $5$ and $-1$ satisfy the original equation, $ \sqrt{2x + 6} + 4 = x + 3 $. When $x = -1$, the original equation gives $ \sqrt{2(-1) + 6} + 4 = (-1) + 3 $, or $6 = 2$, which is false. Therefore, $-1$ does not satisfy the original equation. When $x = 5$, the original equation gives $ \sqrt{2(5) + 6} + 4 = 5 + 3 $, or $8 = 8$, which is true. Therefore, $x = 5$ is the only solution to the original equation, and so the solution set is $ \{5\} $. Choices A, C, and D are incorrect because each of these sets contains at least one value that results in a false statement when substituted into the given equation. For instance, in choice D, when 0 is substituted for $x$ into the given equation, the result is $ \sqrt{2(0)} + 6 + 4 = (0) + 3 $, or $ \sqrt{6} + 4 = 3 $. This is not a true statement, so 0 is not a solution to the given equation.

If \sqrt{x+3} + \sqrt{x-1} = 4, what is the value of x?

C. \frac{13}{4}. To solve the equation $\sqrt{x+3} + \sqrt{x-1} = 4$, we first isolate one of the square roots: $\sqrt{x+3} = 4 - \sqrt{x-1}$. Squaring both sides gives $x+3 = (4 - \sqrt{x-1})^2$. Expanding the right side, we have $x+3 = 16 - 8\sqrt{x-1} + (x-1)$. Simplifying, $x+3 = x + 15 - 8\sqrt{x-1}$. This reduces to $8\sqrt{x-1} = 12$, so $\sqrt{x-1} = \frac{3}{2}$. Squaring again, $x-1 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$. Thus, $x = \frac{9}{4} + 1 = \frac{13}{4}$.

What value of x satisfies the equation \sqrt{x + 4} = 11?

C. 117. The correct answer is 117. Squaring both sides of the given equation gives $x + 4 = 11^2$, or $x + 4 = 121$. Subtracting 4 from both sides of this equation gives $x = 117$.

SAT Math skill page

SAT Radical Equations Practice

Isolate the radical before squaring, then check every candidate in the original equation.

Practice radical equations in Dolphin SAT Practice questions

12-18 min practice time3 questions on pageAdvanced Math

Practice time12-18 min

Question bank3 questions

Best forAdvanced Math

What this tests

What to know for this SAT skill

Isolate a square-root expression before squaring
Solve equations created by removing a radical
Recognize when squaring introduces an extra solution
Check candidate solutions in the original equation

Dolphin question bank

Practice your skill

Question 1 of 3Hard

What is the solution set of the equation

2 x + 6 + 4 = x + 3

{-1}
{5}
{-1, 5}
{0, -1, 5}

Show answer and solution

Correct answerB. {5}

Solution walkthrough

Subtracting 4 from both sides of $2 x + 6 + 4 = x + 3$ isolates the radical expression on the left side of the equation as follows: $2 x + 6 = x - 1$ . Squaring both sides of $2 x + 6 = x - 1$ yields $2 x + 6 = x^{2} - 2 x + 1$ . This equation can be rewritten as a quadratic equation in standard form: $x^{2} - 4 x - 5 = 0$ .
One way to solve this quadratic equation is to factor the expression $x^{2} - 4 x - 5$ by identifying two numbers with a sum of $- 4$ and a product of $- 5$ . These numbers are $- 5$ and $1$ . So the quadratic equation can be factored as $(x - 5) (x + 1) = 0$ .
It follows that $5$ and $- 1$ are the solutions to the quadratic equation. However, the solutions must be verified by checking whether $5$ and $- 1$ satisfy the original equation, $2 x + 6 + 4 = x + 3$ . When $x = - 1$ , the original equation gives $2 (- 1) + 6 + 4 = (- 1) + 3$ , or $6 = 2$ , which is false.
Therefore, $- 1$ does not satisfy the original equation. When $x = 5$ , the original equation gives $2 (5) + 6 + 4 = 5 + 3$ , or $8 = 8$ , which is true. Therefore, $x = 5$ is the only solution to the original equation, and so the solution set is ${5}$ .

Why the other choices miss

Choices A, C, and D are incorrect because each of these sets contains at least one value that results in a false statement when substituted into the given equation. For instance, in choice D, when 0 is substituted for

x

into the given equation, the result is

2 (0) + 6 + 4 = (0) + 3

, or

6 + 4 = 3

. This is not a true statement, so 0 is not a solution to the given equation.

Question 2 of 3Medium

x + 3 + x - 1 = 4

, what is the value of x?

2
3
$\frac{13}{4}$
4

Show answer and solution

Correct answer $C . \frac{13}{4}$

Solution walkthrough

To solve the equation $x + 3 + x - 1 = 4$ , we first isolate one of the square roots: $x + 3 = 4 - x - 1$ . Squaring both sides gives $x + 3 = (4 - x - 1)^{2}$ .
Expanding the right side, we have $x + 3 = 16 - 8 x - 1 + (x - 1)$ . Simplifying, $x + 3 = x + 15 - 8 x - 1$ .
This reduces to $8 x - 1 = 12$ , so $x - 1 = \frac{3}{2}$ . Squaring again, $x - 1 = (\frac{3}{2})^{2} = \frac{9}{4}$ .
Thus, $x = \frac{9}{4} + 1 = \frac{13}{4}$ .

Question 3 of 3Easy

What value of x satisfies the equation

x + 4 = 11

Show answer and solution

Correct answerC. 117

Solution walkthrough

The correct answer is 117.
Squaring both sides of the given equation gives $x + 4 = 1 1^{2}$ , or $x + 4 = 121$ .
Subtracting 4 from both sides of this equation gives $x = 117$ .

Avoid these traps

Common mistakes on this skill

Squaring before isolating the radical

Move all non-radical terms away from the root first so the algebra stays manageable.

Forgetting to square an entire side

If a side contains multiple terms, square the complete expression rather than each term separately.

Keeping an extraneous solution

Squaring can create a candidate that does not satisfy the original equation, so every result must be checked.

Study plan

How to practice this skill in Dolphin

Isolate the radical expression.
Square both complete sides of the equation.
Solve the resulting equation and list every candidate.
Substitute each candidate into the original radical equation.

Practice radical equations in Dolphin SAT

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FAQ

Questions about SAT Radical Equations Practice

Why can squaring create extraneous solutions?

Positive and negative values can have the same square, so squaring may erase a sign condition from the original equation.

Does the SAT test radical equations without a calculator?

Yes. The arithmetic is usually manageable when you isolate the radical and check the result systematically.

What should I check before solving?

A principal square root is nonnegative, so the expression on the other side must also be nonnegative.