ACCUPLACER Advanced Algebra & Functions Quiz: Solving Radical Equations

18 questions · exam conditions

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Solving Radical EquationsQuestion 1 of 18

What is the sum of all real solutions to the equation $\sqrt{3x + 10} = x$ ?

-2

ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Solving Radical Equations

Practice Solving Radical Equations in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Radical Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the sum of all real solutions to the equation $\sqrt{3x + 10} = x$ ?

-2
3
5 (correct answer)
15

Explanation: Square both sides of the equation to get

3x + 10 = x^2

. Rearrange this into a standard quadratic form:

x^2 - 3x - 10 = 0

. Factor the quadratic:

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

. The potential solutions are

x = 5

and

x = -2

. Check these in the original equation. For

x = 5

\sqrt{3(5) + 10} = 5 \Rightarrow \sqrt{25} = 5 \Rightarrow 5 = 5

, which is correct. For

x = -2

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

, which is incorrect because the principal square root must be non-negative. Therefore,

x = -2

is an extraneous solution. The only real solution is 5, so the sum of the solutions is 5.

Question 2

Solve for $y$ : $3\sqrt{y} = \sqrt{4y + 5}$ .

-1
No solution
5
1 (correct answer)

Explanation: To solve this equation, square both sides:

(3\sqrt{y})^2 = (\sqrt{4y + 5})^2

. This simplifies to

9y = 4y + 5

. Next, solve the resulting linear equation for

y

. Subtract

4y

from both sides:

5y = 5

. Divide by 5:

y = 1

. Finally, check the solution in the original equation:

3\sqrt{1} = \sqrt{4(1) + 5} \Rightarrow 3(1) = \sqrt{9} \Rightarrow 3 = 3

. The solution is valid.

Question 3

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

1
4 (correct answer)
9
No solution

Explanation: First, isolate one of the radical terms. Let's move

\sqrt{x}

to the other side:

\sqrt{x - 3} = 3 - \sqrt{x}

. Now, square both sides:

(\sqrt{x - 3})^2 = (3 - \sqrt{x})^2

, which gives

x - 3 = 9 - 6\sqrt{x} + x

. Subtract

x

from both sides:

-3 = 9 - 6\sqrt{x}

. Isolate the remaining radical term:

-12 = -6\sqrt{x}

, which simplifies to

2 = \sqrt{x}

. Square both sides again to solve for

x

4 = x

. Finally, check the solution in the original equation:

\sqrt{4 - 3} + \sqrt{4} = 3 \Rightarrow \sqrt{1} + 2 = 3 \Rightarrow 1 + 2 = 3

. The solution is correct.

Question 4

What is the solution set for the equation $2 + \sqrt[3]{5x - 2} = 5$ ?

$x = 1$
$x = \frac{11}{5}$
$x = \frac{29}{5}$ (correct answer)
$x = 25$

Explanation: First, isolate the radical term by subtracting 2 from both sides:

\sqrt[3]{5x - 2} = 3

. To eliminate the cube root, cube both sides of the equation:

(\sqrt[3]{5x - 2})^3 = 3^3

. This simplifies to

5x - 2 = 27

. Add 2 to both sides:

5x = 29

. Divide by 5 to solve for

x

x = \frac{29}{5}

. Odd-indexed roots do not typically introduce extraneous solutions, so checking is mainly for catching calculation errors.

Question 5

Which of the following equations has no real solutions?

$\sqrt{x + 1} = x - 3$ (correct answer)
$\sqrt{x - 2} = x - 4$
$\sqrt{x + 3} = x - 1$
$\sqrt{x - 4} = x - 6$

Explanation: For a radical equation

\sqrt{f(x)} = g(x)

to have real solutions, we need

f(x) \geq 0

and

g(x) \geq 0

, and the equation must check out after squaring. For choice A:

\sqrt{x + 1} = x - 3

requires

x \geq -1

and

x \geq 3

, so

x \geq 3

. Squaring:

x + 1 = (x - 3)^2 = x^2 - 6x + 9

, so

x^2 - 7x + 8 = 0

. This gives

x = \frac{7 \pm \sqrt{49 - 32}}{2} = \frac{7 \pm \sqrt{17}}{2}

. Both values are less than 3, violating our domain restriction. For choice B: Similar analysis shows this has valid solutions. Choice C and D also yield valid solutions when properly checked. Only choice A has domain restrictions that eliminate all algebraic solutions.

Question 6

Which of the following describes the solution(s) to $\sqrt[3]{x^2 - 7} = 3$ ?

$x = 4$ or $x = -4$
$x = 6$ or $x = -6$
$x = \sqrt{34}$ or $x = -\sqrt{34}$ (correct answer)
$x = 2\sqrt{5}$ or $x = -2\sqrt{5}$

Explanation: To solve the equation, cube both sides to eliminate the cube root:

(\sqrt[3]{x^2 - 7})^3 = 3^3

. This simplifies to

x^2 - 7 = 27

. Add 7 to both sides to isolate the

x^2

term:

x^2 = 34

. Finally, take the square root of both sides, remembering to include both the positive and negative roots:

x = \pm\sqrt{34}

. So the solutions are

x = \sqrt{34}

and

x = -\sqrt{34}

Question 7

If $(2x - 1)^{1/2} - x = -2$ , what is the value of $x$ ?

1
No solution
1 and 5
5 (correct answer)

Explanation: Rewrite the equation using a radical:

\sqrt{2x - 1} - x = -2

. Isolate the radical term:

\sqrt{2x - 1} = x - 2

. Square both sides:

2x - 1 = (x - 2)^2

. Expand the right side:

2x - 1 = x^2 - 4x + 4

. Rearrange into a quadratic equation:

x^2 - 6x + 5 = 0

. Factor the quadratic:

(x - 1)(x - 5) = 0

. The potential solutions are

x = 1

and

x = 5

. Check these solutions. For

x = 1

\sqrt{2(1) - 1} - 1 = -2 \Rightarrow \sqrt{1} - 1 = -2 \Rightarrow 1 - 1 = -2 \Rightarrow 0 = -2

, which is false. For

x = 5

\sqrt{2(5) - 1} - 5 = -2 \Rightarrow \sqrt{9} - 5 = -2 \Rightarrow 3 - 5 = -2

, which is true. The only solution is

x=5

Question 8

What is the value of $x$ that satisfies the equation $\sqrt{4x^2 + 5x - 10} = 2x$ ?

2 (correct answer)
-2
5
No solution

Explanation: First, note that for the radical to be defined and equal to

2x

, we must have

2x \ge 0

, so

x \ge 0

. Square both sides of the equation:

4x^2 + 5x - 10 = (2x)^2

. This simplifies to

4x^2 + 5x - 10 = 4x^2

. Subtract

4x^2

from both sides, leaving a linear equation:

5x - 10 = 0

. Solve for

x

5x = 10 \Rightarrow x = 2

. This solution satisfies the condition

x \ge 0

. Check it in the original equation:

\sqrt{4(2^2) + 5(2) - 10} = 2(2) \Rightarrow \sqrt{16 + 10 - 10} = 4 \Rightarrow \sqrt{16} = 4

, which is true. The solution is

x = 2

Question 9

The equation $x + \sqrt{x+1} = 5$ has one solution. What is this solution?

3 (correct answer)
8
24
No solution

Explanation: Isolate the radical term by subtracting

x

from both sides:

\sqrt{x+1} = 5 - x

. Square both sides:

x+1 = (5 - x)^2

. Expand the right side:

x+1 = 25 - 10x + x^2

. Rearrange into a quadratic equation set to zero:

x^2 - 11x + 24 = 0

. Factor the quadratic:

(x - 3)(x - 8) = 0

. The potential solutions are

x = 3

and

x = 8

. Check these in the original equation. For

x = 3

3 + \sqrt{3+1} = 5 \Rightarrow 3 + \sqrt{4} = 5 \Rightarrow 3 + 2 = 5

, which is true. For

x = 8

8 + \sqrt{8+1} = 5 \Rightarrow 8 + \sqrt{9} = 5 \Rightarrow 8 + 3 = 5 \Rightarrow 11 = 5

, which is false. The only valid solution is

x = 3

Question 10

Solve for $x$ : $\sqrt{x} = \sqrt{x-9} + 1$

25 (correct answer)
16
100
No solution

Explanation: Square both sides of the equation:

(\sqrt{x})^2 = (\sqrt{x-9} + 1)^2

. This gives

x = (x-9) + 2\sqrt{x-9} + 1

. Simplify the right side:

x = x - 8 + 2\sqrt{x-9}

. Subtract

x

from both sides and add 8 to both sides to isolate the radical term:

8 = 2\sqrt{x-9}

. Divide by 2:

4 = \sqrt{x-9}

. Square both sides again:

16 = x - 9

. Solve for

x

x = 25

. Check the solution in the original equation:

\sqrt{25} = \sqrt{25-9} + 1 \Rightarrow 5 = \sqrt{16} + 1 \Rightarrow 5 = 4 + 1

. The solution is correct.

Question 11

Solve the equation $x = \sqrt{3x - 2} + 2$ .

$x = 1$
$x = 6$ (correct answer)
$x = 1$ and $x = 6$
No solution

Explanation: First, isolate the radical by subtracting 2 from both sides:

x - 2 = \sqrt{3x - 2}

. Square both sides:

(x - 2)^2 = 3x - 2

. Expand the left side:

x^2 - 4x + 4 = 3x - 2

. Move all terms to one side to form a quadratic equation:

x^2 - 7x + 6 = 0

. Factor the quadratic:

(x - 1)(x - 6) = 0

. The potential solutions are

x = 1

and

x = 6

. Check both in the original equation. For

x = 1

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

, which is false. For

x = 6

6 = \sqrt{3(6) - 2} + 2 \Rightarrow 6 = \sqrt{16} + 2 \Rightarrow 6 = 4 + 2

, which is true. The only valid solution is

x = 6

Question 12

What is the sum of the solutions for the equation $\sqrt{2x + 1} - \sqrt{x} = 1$ ?

0
2
4 (correct answer)
5

Explanation: First, isolate one radical:

\sqrt{2x + 1} = 1 + \sqrt{x}

. Square both sides:

2x + 1 = (1 + \sqrt{x})^2

, which expands to

2x + 1 = 1 + 2\sqrt{x} + x

. Simplify and isolate the remaining radical term:

x = 2\sqrt{x}

. Square both sides again:

x^2 = (2\sqrt{x})^2

, which gives

x^2 = 4x

. Rearrange into

x^2 - 4x = 0

and factor:

x(x - 4) = 0

. The potential solutions are

x = 0

and

x = 4

. Check both solutions in the original equation. For

x = 0

\sqrt{2(0)+1} - \sqrt{0} = 1 \Rightarrow 1 - 0 = 1

, which is true. For

x = 4

\sqrt{2(4)+1} - \sqrt{4} = 1 \Rightarrow \sqrt{9} - 2 = 1 \Rightarrow 3 - 2 = 1

, which is true. Both solutions are valid. The sum of the solutions is

0 + 4 = 4

Question 13

Find the solution set for $\sqrt{x^2 - 3x + 3} = 1$ .

$\{-1, -2\}$
$\{1\}$
$\{2\}$
$\{1, 2\}$ (correct answer)

Explanation: To solve, square both sides of the equation:

x^2 - 3x + 3 = 1^2

, which simplifies to

x^2 - 3x + 3 = 1

. Set the equation to zero:

x^2 - 3x + 2 = 0

. Factor the quadratic expression:

(x - 1)(x - 2) = 0

. This gives two potential solutions:

x = 1

and

x = 2

. Checking both solutions in the original equation is necessary. For

x = 1

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

. This is true. For

x = 2

\sqrt{2^2 - 3(2) + 3} = \sqrt{4 - 6 + 3} = \sqrt{1} = 1

. This is also true. Both solutions are valid.

Question 14

What is the solution set for the equation $\sqrt{x + 7} = x - 5$ ?

$\{2\}$
$\{9\}$ (correct answer)
$\{2, 9\}$
$\{18\}$

Explanation: To solve the equation, first square both sides to eliminate the radical:

x + 7 = (x - 5)^2

. Expanding the right side gives

x + 7 = x^2 - 10x + 25

. Rearrange the terms to form a quadratic equation:

x^2 - 11x + 18 = 0

. Factoring the quadratic yields

(x - 2)(x - 9) = 0

, which gives potential solutions

x = 2

and

x = 9

. Each potential solution must be checked in the original equation. For

x = 9

\sqrt{9 + 7} = 9 - 5 \Rightarrow \sqrt{16} = 4 \Rightarrow 4 = 4

, which is true. For

x = 2

\sqrt{2 + 7} = 2 - 5 \Rightarrow \sqrt{9} = -3 \Rightarrow 3 = -3

, which is false. Thus,

x = 2

is an extraneous solution. The only valid solution is

x = 9

Question 15

Solve for $x$ : $2\sqrt{x + 1} - x = -2$

$x = 0$
$x = 8$ (correct answer)
$x = 0$ and $x = 8$
No solution

Explanation: First, isolate the radical term:

2\sqrt{x + 1} = x - 2

. Square both sides of the equation:

(2\sqrt{x + 1})^2 = (x - 2)^2

, which simplifies to

4(x + 1) = x^2 - 4x + 4

. Distribute on the left side:

4x + 4 = x^2 - 4x + 4

. Set the equation to zero:

0 = x^2 - 8x

. Factor out

x

0 = x(x - 8)

. The potential solutions are

x = 0

and

x = 8

. Check each solution in the original equation. For

x = 0

2\sqrt{0 + 1} - 0 = -2 \Rightarrow 2(1) = -2 \Rightarrow 2 = -2

, which is false. For

x = 8

2\sqrt{8 + 1} - 8 = -2 \Rightarrow 2\sqrt{9} - 8 = -2 \Rightarrow 2(3) - 8 = -2 \Rightarrow 6 - 8 = -2

, which is true. The only solution is

x = 8

Question 16

Which value of $x$ satisfies $\sqrt{x^2 - 8} = x - 2$ ?

$x = 2$ only
$x = 3$ only (correct answer)
$x = -2$ only
$x = 3$ and $x = -2$

Explanation: Square both sides:

x^2 - 8 = (x - 2)^2 = x^2 - 4x + 4

. Simplifying:

-8 = -4x + 4

, so

4x = 12

and

x = 3

. We must check this solution: For

x = 3

\sqrt{9 - 8} = \sqrt{1} = 1

and

3 - 2 = 1

✓. We must also check the domain:

x^2 - 8 \geq 0

requires

x \leq -2\sqrt{2}

x \geq 2\sqrt{2} \approx 2.83

, and

x - 2 \geq 0

requires

x \geq 2

. Since

x = 3

satisfies both conditions, it's valid. Choice A assumes

x = 2

works without checking. Choice C comes from solving incorrectly and not checking domain restrictions. Choice D assumes both solutions from a quadratic would be valid without verification.

Question 17

How many solutions does the equation $\sqrt{6 - x} = x + 2$ have?

No real solutions
Exactly one solution (correct answer)
Exactly two solutions
Infinitely many solutions

Explanation: Square both sides:

6 - x = (x + 2)^2 = x^2 + 4x + 4

. Rearranging:

x^2 + 5x - 2 = 0

. Using the quadratic formula:

x = \frac{-5 \pm \sqrt{25 + 8}}{2} = \frac{-5 \pm \sqrt{33}}{2}

. This gives two algebraic solutions, but we must check both. For the equation to be valid, we need

6 - x \geq 0

(so

x \leq 6

) and

x + 2 \geq 0

(so

x \geq -2

). The solution

x = \frac{-5 + \sqrt{33}}{2} \approx 0.37

satisfies these conditions and checks out. The solution

x = \frac{-5 - \sqrt{33}}{2} \approx -5.37

violates

x \geq -2

, so it's extraneous. Choice A suggests no solutions exist. Choice C assumes both algebraic solutions are valid. Choice D misunderstands the nature of radical equations.

Question 18

If $\sqrt{x + 4} - \sqrt{x - 1} = 1$ , then $x$ equals:

5 (correct answer)
8
12
20

Explanation: Isolate one radical:

\sqrt{x + 4} = 1 + \sqrt{x - 1}

. Square both sides:

x + 4 = 1 + 2\sqrt{x - 1} + (x - 1) = x + 2\sqrt{x - 1}

. Simplifying:

4 = 2\sqrt{x - 1}

, so

\sqrt{x - 1} = 2

. Square again:

x - 1 = 4

, so

x = 5

. Check:

\sqrt{9} - \sqrt{4} = 3 - 2 = 1

✓. Also verify domain:

x \geq 1

for the second radical. Choice B comes from making an error when squaring the binomial. Choice C results from incorrectly isolating the radical. Choice D represents an arithmetic error in the final steps.