Algebra Quiz: Solving Rational And Radical Equations

20 questions · exam conditions

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Solving Rational And Radical EquationsQuestion 1 of 20

Solve and check for extraneous solutions:

$\frac{x}{x-3}=2$

\{6\}

\{3\}

\{-6\}

\{0\}

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Algebra Quiz

Algebra Quiz: Solving Rational And Radical Equations

Practice Solving Rational And Radical Equations in Algebra 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 Rational And Radical Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

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

Solve and check for extraneous solutions:

$\frac{x}{x-3}=2$

$\{6\}$ (correct answer)
$\{3\}$
$\{-6\}$
$\{0\}$

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

: (1) The denominator is (x-3), so multiply both sides by (x-3):

x = 2(x-3)

. (2) Simplify:

x = 2x - 6

. (3) Solve:

x - 2x = -6

, so

-x = -6

, thus

x = 6

. (4) Check in original: Does x = 6 make the denominator zero?

6 - 3 = 3 ≠ 0

, so it's valid! Verify it satisfies the equation:

\frac{6}{6-3} = \frac{6}{3} = 2

✓. Final answer: {6}. Choice A correctly solves to get x = 6 and verifies it doesn't make the denominator zero, confirming it as the valid solution. Choice B might result from an arithmetic error when solving -x = -6, incorrectly getting x = 3, but x = 3 would make the denominator (x-3) = 0, creating division by zero—definitely extraneous! The rational equation solving recipe: (1) Find the LCD of all denominators, (2) Multiply EVERY term (both sides, all terms) by the LCD—fractions will cancel, (3) Solve the resulting polynomial equation, (4) Check each solution: does it make any original denominator equal zero? If yes, it's extraneous—reject it! If no, verify it satisfies the original equation. Keep only valid solutions. The checking step is non-negotiable!

Question 2

Solve and check for extraneous solutions: $\frac{x}{x-3}=2.$

$\{6,3\}$
$\{6\}$ (correct answer)
$\{0\}$
$\{3\}$

(x - 3)

and our solution is

x = 3

, it's extraneous because it makes the original denominators zero (undefined!). Always check that solutions don't make any denominator in the original equation equal zero. Solving

x/(x-3) = 2

: (1) Cross multiply:

x = 2(x-3)

. (2) Expand:

x = 2x - 6

. (3) Solve:

-x = -6 \to x = 6

. (4) Check in original: Does

x = 6

make any denominator zero? The only denominator is

(x-3)

, and

6-3 = 3 \neq 0

. Good! Verify it satisfies equation:

6/(6-3) = 6/3 = 2

✓. Final answer:

x = 6

. Choice A correctly solves to get

x = 6

with proper checking. Choice B would give

x = 3

, but checking: this makes the denominator

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

, which is undefined! This would be extraneous—we can't have division by zero. Always verify: for rational equations, does the solution make any denominator zero? If yes, reject it as extraneous. The rational equation solving recipe: (1) Find the LCD of all denominators, (2) Multiply EVERY term (both sides, all terms) by the LCD—fractions will cancel, (3) Solve the resulting polynomial equation, (4) Check each solution: does it make any original denominator equal zero? If yes, it's extraneous—reject it! If no, verify it satisfies the original equation. Keep only valid solutions. The checking step is non-negotiable!

Question 3

Solve and verify your answer: $\sqrt{3x+1}=x+1.$

$\{0,1\}$ (both valid) (correct answer)
$\{1\}$ (and $x=0$ is extraneous)
$\{0\}$ (and $x=1$ is extraneous)
$\varnothing$

\sqrt{x + 3} = 5

), we isolate the radical and then square both sides to eliminate the radical, giving a polynomial equation. But squaring can introduce extraneous solutions: if the original has

\sqrt{...} = \text{negative}

, there's no real solution, but squaring gives a positive equation that might have solutions. Always substitute back into the ORIGINAL equation to verify—if it doesn't work, it's extraneous! Solving

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

: (1) Square both sides:

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

→

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

. (2) Simplify:

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

→

3x = x^2 + 2x

→

x = x^2

. (3) Rearrange:

x^2 - x = 0

→

x(x-1) = 0

. (4) Solve:

x = 0

x = 1

. (5) Check each in ORIGINAL equation: For

x = 0

\sqrt{3(0)+1} = \sqrt{1} = 1

, and

0+1 = 1

✓ Valid! For

x = 1

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

, and

1+1 = 2

✓ Valid! Therefore, both

x = 0

and

x = 1

are valid solutions. Choice A correctly identifies both solutions as valid with proper checking. Choice B would reject

x = 0

as extraneous, but checking:

\sqrt{1} = 1

and

0+1 = 1

, so both sides equal 1. This one checks out! Just because a solution is zero doesn't automatically make it extraneous—you must verify by substitution. For radical equations: (1) Isolate the radical on one side if possible (makes squaring cleaner), (2) Square both sides (if square root) or cube (if cube root), (3) Solve the resulting equation (might be quadratic!), (4) Check EVERY solution in the ORIGINAL equation—substitute and see if both sides match. If a solution makes the radical expression negative or doesn't satisfy the equation, it's extraneous. Quick verification trick: when checking a radical equation solution, write out the substitution completely: 'Left side:

\sqrt{[\text{value} + \text{number}]} = \sqrt[\text{result}] = [\text{principal root}]

. Right side:

[\text{calculation}]

. Do they match?' This systematic left-side vs right-side check prevents errors.

Question 4

Solve and check for extraneous solutions:

$\frac{x}{x-5}=3$

$x=\frac{15}{2}$ (correct answer)
$x=5$ (valid)
$x=\frac{5}{2}$
No solution (extraneous)

(x - 3)

and our solution is

x = 3

, it's extraneous because it makes the original denominators zero (undefined!). Always check that solutions don't make any denominator in the original equation equal zero. Solving

\frac{x}{x-5} = 3

: (1) The LCD is

(x-5)

. (2) Multiply both sides by

(x-5)

x = 3(x-5)

. (3) Simplify:

x = 3x - 15

. (4) Solve:

x - 3x = -15 \to -2x = -15 \to x = \frac{15}{2}

. (5) Check in original: Does

x = \frac{15}{2}

make the denominator zero?

x - 5 = \frac{15}{2} - 5 = \frac{15}{2} - \frac{10}{2} = \frac{5}{2} \neq 0

. Good! Now verify it satisfies the equation:

\frac{\frac{15}{2}}{5/2} = \frac{15}{2} \times \frac{2}{5} = \frac{15}{5} = 3

✓. Final answer:

x = \frac{15}{2}

. Choice A correctly solves to get

x = \frac{15}{2}

and this value doesn't make the denominator zero, so it's valid. Choice B suggests

x = 5

is valid, but this would make the denominator

x - 5 = 5 - 5 = 0

, causing division by zero in the original equation. This is the classic extraneous solution for rational equations! Always verify: For rational equations, does the solution make any denominator zero? If yes, it's extraneous and must be rejected. The rational equation solving recipe: (1) Find the LCD of all denominators, (2) Multiply EVERY term (both sides, all terms) by the LCD—fractions will cancel, (3) Solve the resulting polynomial equation, (4) Check each solution: does it make any original denominator equal zero? If yes, it's extraneous—reject it! If no, verify it satisfies the original equation. Keep only valid solutions. The checking step is non-negotiable!

Question 5

Solve and check for extraneous solutions:

$\frac{x}{x-1}=2$

What is the solution set (excluding extraneous solutions)?

$\{2\}$ (correct answer)
$\{1\}$
$\{0\}$
$\varnothing$

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

: (1) The equation is already in a simple form. (2) Cross-multiply:

x = 2(x-1)

. (3) Expand:

x = 2x - 2

. (4) Solve:

x - 2x = -2

, so

-x = -2

, thus

x = 2

. (5) Check in original: Does x = 2 make any denominator zero? The denominator is x - 1, so we check: 2 - 1 = 1 ≠ 0. Good! Now verify it satisfies the equation:

\frac{2}{2-1} = \frac{2}{1} = 2

. ✓ Final answer: {2}. Choice A correctly identifies x = 2 as the only solution, with proper checking that it doesn't make the denominator zero. Choice B would suggest x = 1, but checking: x = 1 makes the denominator 1 - 1 = 0, which is undefined! This is exactly the type of extraneous solution we must reject—even if it somehow emerged from algebraic manipulation, any value that makes a denominator zero must be excluded. The rational equation solving recipe: (1) Identify all denominators and their zeros, (2) Solve the equation algebraically, (3) Check each solution: does it make any original denominator equal zero? If yes, it's extraneous—reject it! If no, verify it satisfies the original equation. Keep only valid solutions. The checking step is non-negotiable!

Question 6

A student is solving $\frac{3x}{x-1} - \frac{2x}{x+1} = \frac{x^2 + 5}{x^2 - 1}$ and multiplies everything by the LCD $(x^2 - 1)$ . After simplification, the student gets $3x(x+1) - 2x(x-1) = x^2 + 5$ . What equation results after expanding and collecting like terms?

$x^2 + 5x = x^2 + 5$ , which simplifies to $5x = 5$ (correct answer)
$x^2 - 5x = x^2 + 5$ , which simplifies to $-5x = 5$
$5x^2 + x = x^2 + 5$ , which simplifies to $4x^2 + x - 5 = 0$
$3x^2 + x = x^2 + 5$ , which simplifies to $2x^2 + x - 5 = 0$

Explanation: After multiplying by the LCD, we have

3x(x+1) - 2x(x-1) = x^2 + 5

. Expanding the left side:

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

and

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

. So the left side becomes

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

. Therefore we have

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

. Subtracting

x^2

from both sides gives

5x = 5

, so

x = 1

. However,

x = 1

makes the original equation undefined, so there is no solution to the original equation.

Question 7

When solving $\sqrt{2x + 3} = x - 3$ , a student obtains the quadratic equation $x^2 - 8x + 6 = 0$ after squaring both sides and simplifying. Using the quadratic formula, the solutions are $x = 4 + \sqrt{10}$ and $x = 4 - \sqrt{10}$ . Which of these solutions, if any, are valid for the original equation?

Both solutions are valid since they both make the radicand $2x + 3$ positive
Only $x = 4 + \sqrt{10}$ is valid since $x = 4 - \sqrt{10}$ makes $x - 3$ negative while the square root is positive (correct answer)
Only $x = 4 - \sqrt{10}$ is valid since $x = 4 + \sqrt{10}$ leads to an incorrect equation when substituted back
Neither solution is valid since both lead to contradictions when checked in the original equation

Explanation: For a solution to be valid in

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

, we need: (1)

2x + 3 ≥ 0

so the square root is defined, and (2)

x - 3 ≥ 0

since the square root is always non-negative. Since

\sqrt{10} ≈ 3.16

, we have

x = 4 + \sqrt{10} ≈ 7.16

and

x = 4 - \sqrt{10} ≈ 0.84

. For

x = 4 + \sqrt{10}

x - 3 = 1 + \sqrt{10} > 0

✓. For

x = 4 - \sqrt{10}

x - 3 = 1 - \sqrt{10} < 0

✗. Since square roots are always non-negative, we cannot have

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

when

x - 3 < 0

. Therefore, only

x = 4 + \sqrt{10}

is valid. Choice A ignores the sign constraint. Choice C reverses which solution is valid. Choice D incorrectly rejects the valid solution.

Question 8

A student attempts to solve $\sqrt{2x - 5} = x - 4$ by squaring both sides immediately. Which statement best describes what the student should do after finding potential solutions?

Check each solution in the original equation since squaring can introduce extraneous solutions (correct answer)
Check each solution in the squared equation since the original equation might have been incorrect
Accept all solutions since squaring both sides preserves the solution set completely
Reject any negative solutions since square roots cannot equal negative numbers in this context

Explanation: When solving radical equations by squaring both sides, extraneous solutions can be introduced because squaring is not a reversible operation (if

a = b

, then

a^2 = b^2

, but if

a^2 = b^2

, then

a = ±b

). The student must check each potential solution in the original equation. For this equation, squaring gives

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

, which leads to

x^2 - 10x + 21 = 0

, with solutions

x = 3

and

x = 7

. Checking: when

x = 3

\sqrt{1} = -1

(false), so

x = 3

is extraneous. When

x = 7

\sqrt{9} = 3

(true). Choice B is incorrect because we check against the original equation. Choice C is wrong because squaring can introduce extraneous solutions. Choice D is incorrect because the issue isn't about negative solutions per se.

Question 9

What is the solution set (excluding extraneous) for $\frac{x+1}{x-1}=2?$

$\{3\}$ (correct answer)
$\{1\}$
$\{0\}$
$\{3,1\}$

Explanation: This question tests your ability to solve rational equations and identify extraneous solutions—solutions that emerge from the solving process but don't actually satisfy the original equation. When solving rational equations (equations with variables in denominators), we multiply both sides by the LCD to clear all the fractions, which gives us a polynomial equation to solve. However, this multiplication can introduce extraneous solutions: if the LCD contains a factor like (x - 3) and our solution is x = 3, it's extraneous because it makes the original denominators zero (undefined!). Always check that solutions don't make any denominator in the original equation equal zero. Solving \frac{x+1}{x-1}=2: (1) Identify the LCD: x-1. (2) Multiply every term by the LCD: x+1 = 2(x-1). (3) Simplify (fractions clear out): x+1 = 2x-2. (4) Solve: 3=x. (5) Check in original: Does x=3 make any denominator zero? 3-1=2 ≠0. Verify it satisfies equation: (3+1)/(3-1)=4/2=2. Final answer: {3}. Choice A correctly identifies {3} as the solution set with proper checking. Choice D includes x=3 and x=1 as valid solutions, but checking: x=1 makes denominator 1-1=0, undefined. This is extraneous! Always verify: (1) For rational, does it make any denominator zero? (2) For radical, does substituting back work? If either fails, reject the solution. The rational equation solving recipe: (1) Find the LCD of all denominators, (2) Multiply EVERY term (both sides, all terms) by the LCD—fractions will cancel, (3) Solve the resulting polynomial equation, (4) Check each solution: does it make any original denominator equal zero? If yes, it's extraneous—reject it! If no, verify it satisfies the original equation. Keep only valid solutions. The checking step is non-negotiable!

Question 10

A student solved the radical equation below by squaring both sides and got $x=1$ and $x=6$ .

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

Which statement correctly identifies the valid solution(s)?

Only $x=1$ is valid; $x=6$ is extraneous
Both $x=1$ and $x=6$ are valid
Neither is valid; both are extraneous
Only $x=6$ is valid; $x=1$ is extraneous (correct answer)

\sqrt{x + 3} = 5

), we isolate the radical and then square both sides to eliminate the radical, giving a polynomial equation. But squaring can introduce extraneous solutions: if the original has

\sqrt{...} =

negative, there's no real solution, but squaring gives a positive equation that might have solutions. Always substitute back into the ORIGINAL equation to verify—if it doesn't work, it's extraneous! Testing the proposed solutions in the original equation

\sqrt{x+3}=x-3

: For

x=1

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

, right side

1-3=-2

2 \neq -2

, extraneous because left positive

\neq

right negative. For

x=6

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

, right side

6-3=3

3=3

, valid because both sides equal. Therefore, only

x=6

is valid,

x=1

is extraneous. Choice C correctly rejects

x=1

as extraneous because it doesn't satisfy the original equation (produces opposite signs) with proper checking. Choice B rejects a valid solution:

x=6

actually DOES satisfy the original equation when we check:

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

. Just because a solution looks unusual or emerged from squaring doesn't automatically make it extraneous—you must verify by substitution. This one checks out! For radical equations: (1) Isolate the radical on one side if possible (makes squaring cleaner), (2) Square both sides (if square root) or cube (if cube root), (3) Solve the resulting equation (might be quadratic!), (4) Check EVERY solution in the ORIGINAL equation—substitute and see if both sides match. If a solution makes the radical expression negative or doesn't satisfy the equation, it's extraneous. Some radical equations have no valid solutions—all turned out extraneous!

Question 11

Solve and check for extraneous solutions: $\frac{3}{x-1}+2=5.$

$\{2\}$ (correct answer)
$\{1\}$
$\{4\}$
$\{2,1\}$

(x - 3)

and our solution is

x = 3

, it's extraneous because it makes the original denominators zero (undefined!). Always check that solutions don't make any denominator in the original equation equal zero. Solving

3/(x-1) + 2 = 5

: (1) First, isolate the fraction:

3/(x-1) = 5 - 2 = 3

. (2) Now we have

3/(x-1) = 3

. (3) Cross multiply:

3 = 3(x-1)

. (4) Simplify:

3 = 3x - 3

. (5) Solve:

6 = 3x \rightarrow x = 2

. (6) Check in original: Does

x = 2

make any denominator zero? The only denominator is

(x-1)

, and

2-1 = 1 \neq 0

. Good! Verify it satisfies equation:

3/(2-1) + 2 = 3/1 + 2 = 3 + 2 = 5 \checkmark

. Final answer:

x = 2

. Choice A correctly solves to get

x = 2

with proper checking. Choice B would give

x = 1

, but checking: this makes the denominator

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

, which is undefined! This would be extraneous if it appeared as a solution. Always verify: for rational equations, does the solution make any denominator zero? If yes, reject it as extraneous. The rational equation solving recipe: (1) Find the LCD of all denominators, (2) Multiply EVERY term (both sides, all terms) by the LCD—fractions will cancel, (3) Solve the resulting polynomial equation, (4) Check each solution: does it make any original denominator equal zero? If yes, it's extraneous—reject it! If no, verify it satisfies the original equation. Keep only valid solutions. The checking step is non-negotiable!

Question 12

A student solved the radical equation

$\sqrt{x+2}=x-4$

and obtained $x=7$ and $x=-1$ after squaring. Which are valid solutions (check in the original equation)?

$\{-1,7\}$ (both valid)
$\{7\}$ only (correct answer)
$\{-1\}$ only
$\varnothing$

\sqrt{x+2}=x-4

: For x = 7: Left side:

\sqrt{7+2} = \sqrt{9} = 3

. Right side:

7-4 = 3

. Since 3 = 3, this is valid! For x = -1: Left side:

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

. Right side:

-1-4 = -5

. Since 1 ≠ -5, this is extraneous! Therefore, only x = 7 is a valid solution. Choice B correctly identifies that only x = 7 is valid, recognizing that x = -1 is extraneous because it doesn't satisfy the original equation when checked. Choice A includes x = -1 as a valid solution, but checking:

\sqrt{-1+2} = 1

while

-1-4 = -5

, and 1 ≠ -5. This is extraneous! Always verify: for radical equations, does substituting back work? If it doesn't match, reject the solution. For radical equations: (1) Isolate the radical on one side if possible (makes squaring cleaner), (2) Square both sides (if square root) or cube (if cube root), (3) Solve the resulting equation (might be quadratic!), (4) Check EVERY solution in the ORIGINAL equation—substitute and see if both sides match. If a solution makes the radical expression negative or doesn't satisfy the equation, it's extraneous. Some radical equations have no valid solutions—all turned out extraneous!

Question 13

When solving $\sqrt{x + 12} - \sqrt{x} = 2$ , a student isolates one radical: $\sqrt{x + 12} = 2 + \sqrt{x}$ . After squaring both sides and simplifying, which equation should result?

$x + 12 = 4 + x + 4\sqrt{x}$
$x + 12 = 4 + x$
$12 = 4 + 4\sqrt{x}$ (correct answer)
$12 = 4\sqrt{x}$

Explanation: When squaring

\sqrt{x + 12} = 2 + \sqrt{x}

, the left side becomes

x + 12

. The right side becomes

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

. So we have

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

. Subtracting

x

from both sides gives

12 = 4 + 4\sqrt{x}

. Choice A shows the equation before simplification. Choice B incorrectly omits the

4\sqrt{x}

term that comes from squaring

(2 + \sqrt{x})^2

. Choice D skips the step of subtracting 4 from both sides.

Question 14

Solve and check for extraneous solutions:

$\frac{3}{x-2}=2$

$x=1$
No solution (extraneous)
$x=\frac{7}{2}$ (correct answer)
$x=\frac{1}{2}$

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

: (1) Multiply both sides by (x-2):

3 = 2(x-2)

. (2) Distribute:

3 = 2x - 4

. (3) Solve:

7 = 2x

, so

x = \frac{7}{2}

. (4) Check in original: Does

x = \frac{7}{2}

make the denominator zero?

\frac{7}{2} - 2 = \frac{3}{2} \neq 0

. Good! Verify it satisfies equation:

\frac{3}{\frac{3}{2}} = \frac{3 \cdot 2}{3} = 2

. ✓ Final answer:

x = \frac{7}{2}

. Choice C correctly solves to get

x = \frac{7}{2}

and verifies it doesn't make the denominator zero, confirming it's a valid solution. Choice A gives

x = 1

, which would make the left side

\frac{3}{1-2} = \frac{3}{-1} = -3

, not 2. This is an arithmetic error—always double-check your algebra! The rational equation solving recipe: (1) Find the LCD of all denominators, (2) Multiply EVERY term (both sides, all terms) by the LCD—fractions will cancel, (3) Solve the resulting polynomial equation, (4) Check each solution: does it make any original denominator equal zero? If yes, it's extraneous—reject it! If no, verify it satisfies the original equation. Keep only valid solutions. The checking step is non-negotiable!

Question 15

Solve and check for extraneous solutions: $\sqrt{x+3}=x-3.$

$\{1,6\}$ (both valid)
$\varnothing$
$\{1\}$ only (and $x=6$ is extraneous)
$\{6\}$ only (and $x=1$ is extraneous) (correct answer)

\sqrt{x + 3} = 5

), we isolate the radical and then square both sides to eliminate the radical, giving a polynomial equation. But squaring can introduce extraneous solutions: if the original has

\sqrt{(\dots)} = \text{negative}

\sqrt{x+3} = x-3

: (1) Isolate radical (already). (2) Square both sides:

x+3 = (x-3)^2

→

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

. (3) Solve:

x^2-7x+6=0

→

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

x=1

6

. (4) Check each in ORIGINAL: For

x=1

\sqrt{4}=2

1-3=-2

2≠-2

, extraneous. For

x=6

\sqrt{9}=3

6-3=3

, matches. Valid solution

x=6

, extraneous

x=1

. Choice B correctly solves to get

\{6\}

only and rejects

x=1

as extraneous because it produces a negative on the right while left is positive, with proper checking. Choice A

\{1,6\}

includes

x=1

as a valid solution, but checking:

\sqrt{1+3}=2 ≠1-3=-2

, doesn't satisfy original. This is extraneous! Always verify: for radical, does substituting back work? If it fails, reject the solution. For radical equations: (1) Isolate the radical on one side if possible (makes squaring cleaner), (2) Square both sides (if square root), (3) Solve the resulting equation (might be quadratic!), (4) Check EVERY solution in the ORIGINAL equation—substitute and see if both sides match. If a solution makes the radical expression negative or doesn't satisfy the equation, it's extraneous. Some radical equations have no valid solutions—all turned out extraneous!

Question 16

Solve and check for extraneous solutions:

$\frac{x}{x-2}=3$

$x=3$ (correct answer)
$x=-3$
$x=2$
No solution (the only solution makes the denominator zero)

x=3

, it's extraneous because it makes the original denominators zero (undefined!). Always check that solutions don't make any denominator in the original equation equal zero. Solving the rational equation

x/(x-2)=3

: (1) Identify the LCD: x-2. (2) Multiply every term by the LCD:

x=3(x-2)

. (3) Simplify:

x=3x-6 \rightarrow -2x=-6 \rightarrow x=3

. (4) Check in original: Does

x=3

make denominator zero?

3-2=1 \neq 0

. Verify it satisfies:

3/1=3

, yes—valid! Final answer:

x=3

. Choice A correctly solves to get

x=3

and confirms it's valid with proper checking, as it doesn't make the denominator zero and satisfies the equation. Choice D states no solution (the only solution makes the denominator zero), but while

x=2

would be extraneous (makes denominator zero), our solution

x=3

is valid—don't assume the root of the denominator is always the solution. The rational equation solving recipe: (1) Find the LCD of all denominators, (2) Multiply EVERY term (both sides, all terms) by the LCD—fractions will cancel, (3) Solve the resulting polynomial equation, (4) Check each solution: does it make any original denominator equal zero? If yes, it's extraneous—reject it! If no, verify it satisfies the original equation. Keep only valid solutions. The checking step is non-negotiable!

Question 17

Solve and check for extraneous solutions: $\sqrt{x+5}+1=x.$

$\{4\}$ (correct answer)
$\{5\}$
$\{4,0\}$
No real solution

Explanation: This question tests your ability to solve radical equations and identify extraneous solutions—solutions that emerge from the solving process but don't actually satisfy the original equation. When solving radical equations (equations with variables under radicals like √(x + 3) = 5), we isolate the radical and then square both sides to eliminate the radical, giving a polynomial equation. But squaring can introduce extraneous solutions: if the original has √(...) = negative, there's no real solution, but squaring gives a positive equation that might have solutions. Always substitute back into the ORIGINAL equation to verify—if it doesn't work, it's extraneous! Solving \sqrt{x+5}+1=x: (1) Isolate radical: √(x+5)=x-1. (2) Square both sides: x+5 = (x-1)^2 → x+5 = x^2-2x+1. (3) Solve: x^2-3x-4=0 → (x-4)(x+1)=0, x=4 or x=-1. (4) Check each in ORIGINAL: For x=4: √(4+5)+1=√9+1=3+1=4, matches x=4, valid. For x=-1: √(-1+5)+1=√4+1=2+1=3 ≠ x=-1, extraneous. Conclusion: Valid solution is 4, extraneous is -1. Choice A correctly identifies {4} as the solution set with proper checking. Choice D says no real solution, but we found a valid one; perhaps from rejecting both without checking, but x=4 works—always verify by substitution! For radical equations: (1) Isolate the radical on one side if possible (makes squaring cleaner), (2) Square both sides (if square root) or cube (if cube root), (3) Solve the resulting equation (might be quadratic!), (4) Check EVERY solution in the ORIGINAL equation—substitute and see if both sides match. If a solution makes the radical expression negative or doesn't satisfy the equation, it's extraneous. Some radical equations have no valid solutions—all turned out extraneous!

Question 18

Solve and check for extraneous solutions: $\sqrt{x} = 2 - x.$

$\varnothing$
$\{1,4\}$ (both valid)
$\{4\}$ only (and $x=1$ is extraneous)
$\{1\}$ only (and $x=4$ is extraneous) (correct answer)

\sqrt{x + 3} = 5

), we isolate the radical and then square both sides to eliminate the radical, giving a polynomial equation. But squaring can introduce extraneous solutions: if the original has

\sqrt{...} = \text{negative}

\sqrt{x} = 2 - x

: (1) Isolate radical (already). (2) Square both sides:

x = (2 - x)^2 \to x = 4 - 4x + x^2

. (3) Solve:

x^2 - 5x + 4 = 0 \to (x - 1)(x - 4) = 0

x = 1

4

. (4) Check each in ORIGINAL: For

x=1

\sqrt{1} = 1

2 - 1 = 1

, matches. For

x=4

\sqrt{4} = 2

2 - 4 = -2

2 \neq -2

, extraneous. Valid

x=1

, extraneous

x=4

. Choice A correctly identifies

\{1\}

only and rejects

x=4

as extraneous because it produces negative on right while left positive, with proper checking. Choice C

\{1,4\}

includes

x=4

as a valid solution, but checking:

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

Question 19

Consider the rational equation $\frac{x+4}{x^2-4} - \frac{2}{x+2} = \frac{1}{x-2}$ . Before solving, a student should recognize that this equation is undefined when:

$x = 2$ only, since this makes the denominator $x-2$ equal to zero
$x = -2$ only, since this makes the denominator $x+2$ equal to zero
$x = 2$ or $x = -2$ , since these values make various denominators equal to zero (correct answer)
$x = 4$ or $x = -4$ , since these values make the first fraction's denominator zero

Explanation: We need to identify all values that make any denominator zero. The first term has denominator

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

, which equals zero when

x = 2

x = -2

. The second term has denominator

x + 2

, which equals zero when

x = -2

. The third term has denominator

x - 2

, which equals zero when

x = 2

. Therefore, the equation is undefined when

x = 2

x = -2

. Choice A misses the restriction from

x^2 - 4

and the term with

x + 2

. Choice B misses the restriction from

x^2 - 4

and the term with

x - 2

. Choice D incorrectly factors

x^2 - 4

Question 20

The equation $\frac{x^2}{x-3} = \frac{9}{x-3}$ appears to have solutions when both sides are multiplied by $(x-3)$ . What is the complete solution set for this equation?

$\{-3, 3\}$ with no restrictions needed
$\{-3\}$ only, since $x = 3$ makes the original equation undefined (correct answer)
$\{3\}$ only, since $x = -3$ is extraneous to the original equation
No solution exists since both potential solutions create undefined expressions

Explanation: Multiplying both sides by

(x-3)

gives

x^2 = 9

, so

x = ±3

. However, we must check these solutions in the original equation. When

x = 3

, we get

\frac{9}{0} = \frac{9}{0}

, which is undefined, so

x = 3

is not a valid solution. When

x = -3

, we get

\frac{9}{-6} = \frac{9}{-6}

, which simplifies to

-\frac{3}{2} = -\frac{3}{2}

, a true statement. Therefore, only

x = -3

is a valid solution. Choice A ignores domain restrictions. Choice C incorrectly identifies which solution is invalid. Choice D incorrectly rejects the valid solution

x = -3