Math 2 Quiz: Solving Radical Equations

11 questions · exam conditions

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Solving Radical EquationsQuestion 1 of 11

If $\sqrt[4]{x + 1} = 3$ , then $\sqrt{x + 1} =$ ?

Math 2 Quiz

Math 2 Quiz: Solving Radical Equations

Practice Solving Radical Equations in Math 2 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 Math 2.

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

If $\sqrt[4]{x + 1} = 3$ , then $\sqrt{x + 1} =$ ?

3
6
9 (correct answer)
12

Explanation: From

\sqrt[4]{x + 1} = 3

, raise both sides to the 4th power:

x + 1 = 3^4 = 81

, so

x = 80

. Now find

\sqrt{x + 1} = \sqrt{80 + 1} = \sqrt{81} = 9

. Alternatively, note that

\sqrt{x + 1} = (\sqrt[4]{x + 1})^2 = 3^2 = 9

. Both methods give the same answer: 9.

Question 2

If $\sqrt{x^2 - 5x + 6} = x - 3$ , then $x$ could equal:

2 or 3
3 only (correct answer)
2 only
1 or 6

Explanation: Square both sides:

x^2 - 5x + 6 = (x - 3)^2 = x^2 - 6x + 9

. Simplifying:

-5x + 6 = -6x + 9

, so

x = 3

. We must check this solution and verify the domain restrictions. For the equation to be valid, we need

x - 3 \geq 0

(since the square root equals

x - 3

), so

x \geq 3

. Also,

x^2 - 5x + 6 \geq 0

. Factoring:

(x - 2)(x - 3) \geq 0

, which holds when

x \leq 2

x \geq 3

. Combining constraints:

x \geq 3

. Since

x = 3

satisfies this, let's verify:

\sqrt{9 - 15 + 6} = \sqrt{0} = 0

and

x - 3 = 0

✓. The only valid solution is

x = 3

Question 3

Consider the equation $x = \sqrt{6x - 5}$ . A student claims that since both sides are positive for valid solutions, squaring will not introduce extraneous solutions. How should you evaluate this claim?

The claim is correct; squaring positive expressions never creates extraneous solutions
The claim is incorrect; the equation has no real solutions due to domain restrictions
The claim is incorrect; squaring can still introduce extraneous solutions even when both sides appear positive (correct answer)
The claim is partially correct; it only applies when the left side is a simple variable

Explanation: The student's claim is incorrect. While it's true that we need both sides to be non-negative before squaring, this doesn't guarantee that squaring won't introduce extraneous solutions. Squaring the equation

x = \sqrt{6x - 5}

gives

x^2 = 6x - 5

, or

x^2 - 6x + 5 = 0

, which factors as

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

. This yields

x = 1

and

x = 5

. However, checking these: For

x = 1

: LHS = 1, RHS =

\sqrt{6(1) - 5} = \sqrt{1} = 1

✓. For

x = 5

: LHS = 5, RHS =

\sqrt{6(5) - 5} = \sqrt{25} = 5

✓. In this case, both solutions are actually valid, but the key point is that the student's reasoning is flawed. Squaring can introduce extraneous solutions regardless of the apparent signs because the squaring operation can make negative expressions appear to satisfy the squared equation when they don't satisfy the original.

Question 4

When solving $\sqrt{x + 4} + \sqrt{x - 1} = 5$ , which of the following represents the most efficient approach to avoid extraneous solutions?

Square both sides immediately, then solve the resulting quadratic equation
Isolate $\sqrt{x + 4}$ , square both sides, simplify, then square again if necessary (correct answer)
Substitute $u = \sqrt{x}$ to convert to a polynomial equation in $u$
Graph both sides and find intersection points using technology

Explanation: Option B is correct because isolating one radical before squaring minimizes the complexity of subsequent algebra and reduces the likelihood of computational errors that can mask extraneous solutions. If we square

\sqrt{x + 4} + \sqrt{x - 1} = 5

immediately, we get

x + 4 + 2\sqrt{(x+4)(x-1)} + x - 1 = 25

, leading to

2\sqrt{x^2 + 3x - 4} = 22 - 2x

, which requires another squaring step anyway. By isolating first:

\sqrt{x + 4} = 5 - \sqrt{x - 1}

, then squaring gives

x + 4 = 25 - 10\sqrt{x - 1} + x - 1

, which simplifies to

10\sqrt{x - 1} = 20

, so

\sqrt{x - 1} = 2

and

x = 5

. Option A creates unnecessary algebraic complexity. Option C doesn't apply since the expressions under the radicals aren't perfect squares of

x

. Option D, while potentially accurate, doesn't demonstrate algebraic problem-solving skills.

Question 5

The equation $\sqrt{x + 7} - \sqrt{x - 2} = 1$ has a solution $x = a$ . What is the value of $\sqrt{a + 7} + \sqrt{a - 2}$ ?

5
7
9 (correct answer)
11

Explanation: Let

u = \sqrt{x + 7}

and

v = \sqrt{x - 2}

. Then

u - v = 1

and we want

u + v

. Note that

u^2 - v^2 = (x + 7) - (x - 2) = 9

. Since

u^2 - v^2 = (u + v)(u - v) = (u + v)(1) = u + v

, we have

u + v = 9

. Therefore,

\sqrt{a + 7} + \sqrt{a - 2} = 9

. To verify this approach works, we can solve for

a

: From

u - v = 1

and

u + v = 9

, we get

u = 5

and

v = 4

. So

\sqrt{x + 7} = 5

gives

x + 7 = 25

, thus

x = 18

. Check:

\sqrt{18 - 2} = \sqrt{16} = 4

✓. The answer is 9.

Question 6

The equation $x = \sqrt{x + 12}$ has how many valid solutions?

0
1 (correct answer)
2
3

Explanation: Square both sides:

x^2 = x + 12

, so

x^2 - x - 12 = 0

. Factor:

(x - 4)(x + 3) = 0

, giving

x = 4

x = -3

. For the original equation to be valid, we need

x \geq 0

(since

\sqrt{x + 12} = x

and square roots are non-negative). Check

x = 4

\sqrt{16} = 4

✓. Check

x = -3

\sqrt{9} = 3

but

x = -3

, so

3 \neq -3

. Therefore,

x = -3

is extraneous. Only

x = 4

is valid, so there is exactly 1 solution.

Question 7

Which of the following is an extraneous solution to the equation $\sqrt{x + 6} = x$ ?

$x = -2$ (correct answer)
$x = 3$
$x = 6$
This equation has no extraneous solutions

Explanation: Square both sides:

x + 6 = x^2

, so

x^2 - x - 6 = 0

. Factor:

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

, giving

x = 3

x = -2

. Check

x = 3

\sqrt{9} = 3

✓. Check

x = -2

\sqrt{4} = 2

but

x = -2

, so

2 \neq -2

. Therefore,

x = -2

is extraneous. The issue is that when we squared both sides, we lost the condition that

\sqrt{x + 6} \geq 0

, but

x

can be negative. For the original equation

\sqrt{x + 6} = x

to hold, we need

x \geq 0

. Since

x = -2 < 0

, it's extraneous.

Question 8

How many solutions does the equation $x + 1 = \sqrt{x^2 + 2x + 5}$ have?

0 (correct answer)
1
2
infinitely many

Explanation: For the equation to be valid, we need

x + 1 \geq 0

, so

x \geq -1

. Square both sides:

(x + 1)^2 = x^2 + 2x + 5

. Expanding:

x^2 + 2x + 1 = x^2 + 2x + 5

. This simplifies to

1 = 5

, which is impossible. Therefore, there are no solutions. The issue is that while

\sqrt{x^2 + 2x + 5} = \sqrt{(x + 1)^2 + 4} \geq 2

for all real

x

, we need

x + 1

to equal this value. But

x + 1 \geq 2

would require

x \geq 1

. Even then,

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

leads to

0 = 4

. The equation has no solutions.

Question 9

For what value of $a$ is $x = 9$ a solution to $\sqrt{x + a} = x - 5$ ?

$a = 25$
$a = 11$
$a = 16$
$a = 7$ (correct answer)

Explanation: When you encounter a radical equation where you need to find a parameter that makes a given value a solution, substitute that value directly into the equation and solve for the unknown parameter. Since

x = 9

must be a solution to

\sqrt{x + a} = x - 5

, substitute

x = 9

into both sides of the equation:

\sqrt{9 + a} = 9 - 5

\sqrt{9 + a} = 4

To solve for

a

, square both sides to eliminate the square root:

9 + a = 16

a = 7

Let's verify: when

a = 7

, the equation becomes

\sqrt{9 + 7} = 9 - 5

, which simplifies to

\sqrt{16} = 4

, or

4 = 4

✓ Now let's see why the other choices don't work. Choice A gives us

a = 25

: substituting yields

\sqrt{9 + 25} = \sqrt{34} \approx 5.83

, but the right side equals

4

. Choice B gives us

a = 11

: this produces

\sqrt{9 + 11} = \sqrt{20} \approx 4.47 \neq 4

. Choice C gives us

a = 16

: this results in

\sqrt{9 + 16} = \sqrt{25} = 5 \neq 4

. The key strategy here is direct substitution. When a problem tells you that a specific value is a solution to an equation containing a parameter, always substitute that value immediately and solve for the parameter. This approach is much more efficient than trying to solve the general equation first.

Question 10

When solving the equation $\sqrt{x + 7} - \sqrt{x - 5} = 2$ , a student obtains potential solutions and must check their validity. Which statement correctly describes the solution process?

The equation has exactly one valid solution: $x = 9$ (correct answer)
The equation has exactly one valid solution: $x = 18$
The equation has two valid solutions: $x = 9$ and $x = 18$
The equation has no valid solutions due to domain restrictions

Explanation: To solve

\sqrt{x + 7} - \sqrt{x - 5} = 2

, we first note the domain requires

x \geq 5

. Isolating one radical:

\sqrt{x + 7} = 2 + \sqrt{x - 5}

. Squaring both sides:

x + 7 = 4 + 4\sqrt{x - 5} + (x - 5)

, which simplifies to

8 = 4\sqrt{x - 5}

\sqrt{x - 5} = 2

. This gives

x - 5 = 4

, so

x = 9

. Checking:

\sqrt{16} - \sqrt{4} = 4 - 2 = 2

✓. The answer is A.

Question 11

A student attempts to solve $\sqrt{2x - 3} = x - 3$ and obtains the solutions $x = 4$ and $x = 6$ . What should the student conclude after checking these solutions?

Both solutions are valid since they satisfy the domain restrictions
Only $x = 6$ is valid; $x = 4$ is an extraneous solution (correct answer)
Only $x = 4$ is valid; $x = 6$ is an extraneous solution
Both solutions are extraneous due to improper algebraic manipulation

Explanation: First, we need

2x - 3 \geq 0

(so

x \geq \frac{3}{2}

) and

x - 3 \geq 0

(so

x \geq 3

) since we're equating a square root to

x - 3

. The domain is

x \geq 3

. Squaring both sides:

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

. Rearranging:

x^2 - 8x + 12 = 0

, which factors as

(x - 4)(x - 6) = 0

. This gives

x = 4

and

x = 6

. Checking

x = 4

\sqrt{2(4) - 3} = \sqrt{5} \approx 2.24

, but

x - 3 = 4 - 3 = 1

. Since

\sqrt{5} \neq 1

x = 4

is extraneous. Checking

x = 6

\sqrt{2(6) - 3} = \sqrt{9} = 3

and

x - 3 = 6 - 3 = 3

. Since

3 = 3

x = 6

is valid.