Math 3 Quiz: Solving Radical Equations

14 questions · exam conditions

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Solving Radical EquationsQuestion 1 of 14

When solving the equation $\sqrt{2x - 3} + \sqrt{x + 1} = 4$ , a student squares both sides to get $2x - 3 + 2\sqrt{(2x-3)(x+1)} + x + 1 = 16$ . What is the correct next step to isolate the remaining radical?

Simplify to get

2\sqrt{(2x-3)(x+1)} = 18 - 3x

, then square both sides

Simplify to get

2\sqrt{(2x-3)(x+1)} = 18 - 3x

, then divide by 2 before squaring

Combine like terms to get

2\sqrt{2x^2 - x - 3} = 18 - 3x

, then square both sides

Factor out 2 to get

\sqrt{(2x-3)(x+1)} = 9 - \frac{3x}{2}

, then square both sides

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Math 3 Quiz

Math 3 Quiz: Solving Radical Equations

Practice Solving Radical Equations in Math 3 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 3.

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

Simplify to get $2\sqrt{(2x-3)(x+1)} = 18 - 3x$ , then square both sides (correct answer)
Simplify to get $2\sqrt{(2x-3)(x+1)} = 18 - 3x$ , then divide by 2 before squaring
Combine like terms to get $2\sqrt{2x^2 - x - 3} = 18 - 3x$ , then square both sides
Factor out 2 to get $\sqrt{(2x-3)(x+1)} = 9 - \frac{3x}{2}$ , then square both sides

Explanation: After expanding

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

, we combine like terms:

3x - 2 + 2\sqrt{(2x-3)(x+1)} = 16

. Isolating the radical term gives

2\sqrt{(2x-3)(x+1)} = 16 - 3x + 2 = 18 - 3x

. The next step is to square both sides. Choice B is incorrect because dividing by 2 is unnecessary at this stage. Choice C incorrectly expands

(2x-3)(x+1)

inside the radical. Choice D makes an error in the algebraic manipulation when factoring.

Question 2

The radical equation $\sqrt{2x - 1} = x - 2$ yields potential solutions when solved algebraically. Without fully solving, what can be determined about the nature of solutions based on domain and range considerations?

Any solution must satisfy $x \geq \frac{1}{2}$ and $x \geq 2$ simultaneously, so $x \geq 2$
The domain requires $x \geq \frac{1}{2}$ but range considerations are irrelevant for radical equations
Solutions must satisfy $x \geq \frac{1}{2}$ from the domain and $x \geq 2$ from range, giving $x \geq 2$ (correct answer)
Domain restrictions eliminate all potential solutions since $\sqrt{2x - 1} \geq 0$ but $x - 2$ can be negative

Explanation: Domain:

2x - 1 \geq 0

requires

x \geq \frac{1}{2}

. Range: Since

\sqrt{2x - 1} \geq 0

, we need

x - 2 \geq 0

, so

x \geq 2

. Both conditions must hold, giving

x \geq 2

. Choice A gets the right answer but with unclear reasoning. Choice B ignores range considerations, which are crucial here. Choice D incorrectly concludes no solutions exist, when the constraints can be simultaneously satisfied.

Question 3

The equation $\sqrt{x + 4} + \sqrt{x - 4} = 4$ can be solved using substitution. If $u = \sqrt{x + 4}$ and $v = \sqrt{x - 4}$ , what additional relationship between $u$ and $v$ can be established?

$u^2 + v^2 = 2x$ , which provides a system of equations with $u + v = 4$
$u^2 - v^2 = 8$ , which combined with $u + v = 4$ allows solving for $u$ and $v$ (correct answer)
$uv = \sqrt{(x+4)(x-4)} = \sqrt{x^2 - 16}$ , creating a system with $u + v = 4$
$(u + v)^2 = u^2 + 2uv + v^2 = 16$ , leading to $2x + 2\sqrt{x^2 - 16} = 16$

Explanation: When solving radical equations with substitution, you're looking for relationships that create a solvable system. Here, with

u = \sqrt{x + 4}

and

v = \sqrt{x - 4}

, you already have

u + v = 4

from the original equation. To find another relationship, consider what you can derive from the definitions of

u

and

v

. Since

u^2 = x + 4

and

v^2 = x - 4

, you can subtract these equations:

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

. This gives you the system

u + v = 4

and

u^2 - v^2 = 8

. Since

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

, you get

8 = 4(u - v)

, so

u - v = 2

. Solving this system yields

u = 3

and

v = 1

, making the correct answer B. Option A states

u^2 + v^2 = 2x

, which is true (since

u^2 + v^2 = x + 4 + x - 4 = 2x

), but this doesn't help solve for

u

and

v

individually since you'd still have three unknowns. Option C gives

uv = \sqrt{x^2 - 16}

, which is correct but creates a more complex system involving nested radicals that's harder to solve than the linear approach in B. Option D expands

(u + v)^2

correctly but leads to the same nested radical complexity as C, making the solution unnecessarily complicated. Study tip: In substitution problems with radicals, look for relationships that eliminate the original variable cleanly—subtracting perfect squares often creates the most manageable system.

Question 4

A radical equation $\sqrt{ax + b} = cx + d$ has the property that when both sides are squared, the resulting quadratic has discriminant equal to zero. What can be concluded about the original radical equation?

The original equation has exactly one solution, and it is guaranteed to be valid
The original equation has exactly one potential solution, which must be checked for validity (correct answer)
The original equation has no real solutions since the discriminant is zero
The original equation has infinitely many solutions due to the zero discriminant

Explanation: When a quadratic has discriminant zero, it has exactly one repeated real root. However, this is a potential solution to the original radical equation. Since squaring can introduce extraneous solutions, this single solution must still be checked in the original equation to confirm validity. It's possible that this solution is extraneous, leaving the original equation with no solutions. Choice A incorrectly guarantees validity. Choice C misinterprets what zero discriminant means. Choice D is incorrect about the number of solutions.

Question 5

A student attempts to solve $x = \sqrt{x + 12}$ by squaring both sides immediately. After obtaining $x^2 = x + 12$ , they find $x = 4$ and $x = -3$ . What is the most important consideration when evaluating these solutions?

Both solutions satisfy the quadratic equation, so both are valid for the original radical equation
The negative solution $x = -3$ must be rejected because $x$ represents the principal square root
Only $x = 4$ is valid because substituting $x = -3$ gives $-3 = \sqrt{9} = 3$ , which is false (correct answer)
The domain restriction $x \geq -12$ eliminates $x = -3$ as a potential solution

Explanation: When checking

x = 4

\sqrt{4 + 12} = \sqrt{16} = 4

, so

x = 4

works. When checking

x = -3

\sqrt{-3 + 12} = \sqrt{9} = 3

, but the original equation requires

x = \sqrt{x + 12}

, so we need

-3 = 3

, which is false. Therefore,

x = -3

is extraneous. Choice A ignores the need to check the original equation. Choice B gives an incorrect reason. Choice D is wrong because

x = -3

satisfies the domain restriction

x \geq -12

, but fails when substituted into the original equation.

Question 6

When solving $\sqrt{2x + 7} + \sqrt{x + 3} = \sqrt{5x + 12}$ , a student squares both sides and obtains $2x + 7 + 2\sqrt{(2x+7)(x+3)} + x + 3 = 5x + 12$ . After simplification, what equation should result?

$\sqrt{(2x+7)(x+3)} = x + 1$ , provided that $x \geq -1$ for the solution to be valid
$\sqrt{2x^2 + 13x + 21} = x + 1$ , which requires checking for extraneous solutions
$2\sqrt{2x^2 + 13x + 21} = 2x + 2$ , which simplifies before squaring again
$\sqrt{(2x+7)(x+3)} = x + 1$ , which can be solved by squaring once more (correct answer)

Explanation: When solving radical equations, you need to isolate radicals systematically and be aware that squaring both sides can introduce extraneous solutions that must be checked later. Starting with the equation after squaring:

2x + 7 + 2\sqrt{(2x+7)(x+3)} + x + 3 = 5x + 12

. Let's simplify the left side by combining like terms:

(2x + x) + (7 + 3) + 2\sqrt{(2x+7)(x+3)} = 3x + 10 + 2\sqrt{(2x+7)(x+3)}

. Setting this equal to the right side:

3x + 10 + 2\sqrt{(2x+7)(x+3)} = 5x + 12

. Subtracting

3x + 10

from both sides gives us:

2\sqrt{(2x+7)(x+3)} = 2x + 2

. Dividing both sides by 2 yields:

\sqrt{(2x+7)(x+3)} = x + 1

. Answer D correctly identifies this result and notes that you can solve it by squaring once more. Answer A incorrectly adds a domain restriction of

x \geq -1

at this stage - while domain considerations matter, this specific restriction isn't the primary focus here. Answer B expands the product under the radical to

\sqrt{2x^2 + 13x + 21}

, which is mathematically equivalent but unnecessarily complicates the expression. Answer C fails to divide both sides by 2, leaving the equation as

2\sqrt{2x^2 + 13x + 21} = 2x + 2

. Remember: when solving radical equations, simplify at each step before proceeding to the next operation. Keep expressions in their most manageable form to avoid algebraic errors.

Question 7

Solve the equation $\sqrt{3x + 7} - \sqrt{x + 3} = 2$ . Which of the following represents the complete solution set?

$x = 6$ (correct answer)
$x = 6$ and $x = -2$
$x = -2$
No real solutions exist

Explanation: To solve

\sqrt{3x + 7} - \sqrt{x + 3} = 2

, isolate one radical:

\sqrt{3x + 7} = 2 + \sqrt{x + 3}

. Square both sides:

3x + 7 = 4 + 4\sqrt{x + 3} + x + 3

, which simplifies to

2x = 4\sqrt{x + 3}

, or

x = 2\sqrt{x + 3}

. Square again:

x^2 = 4(x + 3) = 4x + 12

, so

x^2 - 4x - 12 = 0

. Factoring gives

(x - 6)(x + 2) = 0

, so

x = 6

x = -2

. Checking: For

x = 6

\sqrt{25} - \sqrt{9} = 5 - 3 = 2

✓. For

x = -2

\sqrt{1} - \sqrt{1} = 0 ≠ 2

, so

x = -2

is extraneous. Only

x = 6

is valid.

Question 8

Which equation has the same solution set as $\sqrt{2x - 3} = x - 3$ ?

$2x - 3 = x^2 - 6x + 9$
$2x - 3 = (x - 3)^2$ with $x ≥ 3$
$x^2 - 8x + 12 = 0$ with $x ≥ \frac{3}{2}$
$x^2 - 8x + 12 = 0$ with $x ≥ 3$ (correct answer)

Explanation: Starting with

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

, we need

2x - 3 ≥ 0

(so

x ≥ \frac{3}{2}

) and

x - 3 ≥ 0

(so

x ≥ 3

) since the right side must be non-negative for equality with a square root. The domain is

x ≥ 3

. Squaring both sides:

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

, which gives

x^2 - 8x + 12 = 0

. This factors as

(x - 2)(x - 6) = 0

, so

x = 2

x = 6

. Since we need

x ≥ 3

, only

x = 6

is valid. Checking:

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

and

6 - 3 = 3

✓. Choice A omits the domain restriction. Choice B has the wrong domain. Choice C has the wrong domain restriction.

Question 9

When solving $\sqrt{3x + 1} - \sqrt{2x - 5} = \sqrt{x - 2}$ , a student notes that all three expressions under the radicals must be non-negative. What is the most restrictive domain condition?

$x \geq \frac{5}{2}$ since this is the largest of the three boundary values
$x \geq 2$ since this ensures the right side of the equation is defined
$x \geq -\frac{1}{3}$ since this makes the first radical expression positive
$x \geq \frac{5}{2}$ and the solution must be checked to ensure the left side equals the right side (correct answer)

Explanation: For

\sqrt{3x + 1}

3x + 1 \geq 0 \Rightarrow x \geq -\frac{1}{3}

. For

\sqrt{2x - 5}

2x - 5 \geq 0 \Rightarrow x \geq \frac{5}{2}

. For

\sqrt{x - 2}

x - 2 \geq 0 \Rightarrow x \geq 2

. The most restrictive is

x \geq \frac{5}{2}

since

\frac{5}{2} = 2.5 > 2 > -\frac{1}{3}

. However, domain restrictions alone don't guarantee a solution exists; checking is still required after solving. Choice A gives the correct domain but omits the checking requirement. Choices B and C identify less restrictive conditions.

Question 10

When solving $\sqrt{x^2 - 5x + 6} = x - 3$ , a student finds that $x = 3$ is a potential solution. Upon checking, what issue arises?

The solution is valid since both sides equal zero when $x = 3$ (correct answer)
The solution is invalid because it makes the expression under the radical negative
The solution is invalid because squaring $x - 3$ when $x = 3$ introduces an error
The solution creates a division by zero error when checking the original equation

Explanation: When

x = 3

: Left side:

\sqrt{3^2 - 5(3) + 6} = \sqrt{9 - 15 + 6} = \sqrt{0} = 0

. Right side:

x - 3 = 3 - 3 = 0

. Both sides equal 0, so

x = 3

is indeed a valid solution. Choice B is incorrect because

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

, which equals 0 (not negative) when

x = 3

. Choice C misunderstands the squaring process. Choice D is incorrect as no division by zero occurs.

Question 11

The radical equation $\sqrt{4x + 1} - \sqrt{x - 2} = 3$ requires careful handling of the domain. Before solving, what constraint must be placed on $x$ ?

$x \geq 2$ to ensure both expressions under the radicals are non-negative (correct answer)
$x \geq -\frac{1}{4}$ to ensure the first radical is defined
$x > 2$ to avoid division by zero in the solving process
$x \geq 2$ and $x \neq 3$ to avoid extraneous solutions

Explanation: For

\sqrt{4x + 1}

to be defined, we need

4x + 1 \geq 0

, so

x \geq -\frac{1}{4}

. For

\sqrt{x - 2}

to be defined, we need

x - 2 \geq 0

, so

x \geq 2

. Since both conditions must hold, the domain is

x \geq 2

. Choice B only considers the first radical. Choice C incorrectly suggests strict inequality and mentions division by zero, which isn't relevant here. Choice D adds an unnecessary restriction about avoiding extraneous solutions, which is handled during the checking phase, not domain determination.

Question 12

To solve $\sqrt[4]{3x - 2} = \sqrt{x + 1}$ , a student raises both sides to the fourth power. What is the most efficient alternative approach?

Square both sides twice: first to get $\sqrt{3x - 2} = (x + 1)$ , then square again
Raise both sides to the fourth power directly to get $3x - 2 = (x + 1)^4$
Square both sides to get $\sqrt{3x - 2} = x + 1$ , then square again to get $3x - 2 = (x + 1)^2$ (correct answer)
Convert to exponential form: $(3x - 2)^{1/4} = (x + 1)^{1/2}$ and raise both sides to the fourth power

Explanation: The most efficient approach is to square both sides first:

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

gives

\sqrt{3x-2} = x + 1

. Then square again:

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

, leading to

x^2 - x + 3 = 0

. Choice A has an error in the first step. Choice B works but creates a more complex fourth-degree expansion. Choice D is essentially the same as B but with more notation. Choice C provides the most systematic approach with intermediate steps that are easier to handle.

Question 13

If $\sqrt{x^2 - 6x + 9} + \sqrt{x^2 - 10x + 25} = 2$ , then the solution set is:

$3 ≤ x ≤ 5$ (correct answer)
$x = 3$ or $x = 5$
$x = 4$
No real solutions

Explanation: Notice that

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

and

x^2 - 10x + 25 = (x - 5)^2

. So the equation becomes

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

, which simplifies to

|x - 3| + |x - 5| = 2

. We need to consider different cases based on the signs: Case 1:

x < 3

|x - 3| = 3 - x

and

|x - 5| = 5 - x

, so

(3 - x) + (5 - x) = 8 - 2x = 2

, giving

x = 3

. But

x = 3

is not in the interval

x < 3

. Case 2:

3 ≤ x ≤ 5

|x - 3| = x - 3

and

|x - 5| = 5 - x

, so

(x - 3) + (5 - x) = 2

, which gives

2 = 2

. This is always true for

3 ≤ x ≤ 5

. Case 3:

x > 5

|x - 3| = x - 3

and

|x - 5| = x - 5

, so

(x - 3) + (x - 5) = 2x - 8 = 2

, giving

x = 5

. But

x = 5

is not in the interval

x > 5

. Therefore, the solution is

3 ≤ x ≤ 5

Question 14

The equation $\sqrt{x + 4} + \sqrt{x - 1} = \sqrt{4x + 1}$ has how many valid solutions?

0
1 (correct answer)
2
3

Explanation: Domain requires

x ≥ 1

(most restrictive). Square both sides:

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

. Simplifying:

2x + 3 + 2\sqrt{x^2 + 3x - 4} = 4x + 1

, so

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

, which gives

\sqrt{x^2 + 3x - 4} = x - 1

. For this to be valid, we need

x - 1 ≥ 0

, so

x ≥ 1

(already satisfied). Squaring again:

x^2 + 3x - 4 = (x - 1)^2 = x^2 - 2x + 1

. This simplifies to

5x - 5 = 0

, so

x = 1

. Checking

x = 1

\sqrt{5} + \sqrt{0} = \sqrt{5}

and

\sqrt{5} = \sqrt{5}

✓. Therefore, there is exactly 1 valid solution.