Algebra Quiz: Solving Rational And Radical Equations

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.

Question 1

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

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 - 2) and our solution is x = 2, 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 $\frac{3}{x-2}=2$: (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 2

1. $\{1,6\}$ (both valid)
2. $\varnothing$
3. $\{1\}$ only (and $x=6$ is extraneous)
4. $\{6\}$ only (and $x=1$ is extraneous) (correct answer)

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 $\sqrt{x + 3} = 5$), 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 $\sqrt{(\dots)} = \text{negative}$, 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 $\sqrt{x+3} = x-3$: (1) Isolate radical (already). (2) Square both sides: $x+3 = (x-3)^2$ → $x+3 = x^2-6x+9$. (3) Solve: $x^2-7x+6=0$ → $(x-1)(x-6)=0$, $x=1$ or $6$. (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 3

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

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 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 4

1. $\{4\}$ (correct answer)
2. $\{5\}$
3. $\{4,0\}$
4. 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 5

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

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 $\frac{x}{x-5} = 3$: (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 6

1. $\varnothing$
2. $\{1,4\}$ (both valid)
3. $\{4\}$ only (and $x=1$ is extraneous)
4. $\{1\}$ only (and $x=4$ is extraneous) (correct answer)

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 $\sqrt{x + 3} = 5$), 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 $\sqrt{x} = 2 - x$: (1) Isolate radical (already). (2) Square both sides: $x = (2 - x)^2 \to x = 4 - 4x + x^2$. (3) Solve: $x^2 - 5x + 4 = 0 \to (x - 1)(x - 4) = 0$, $x = 1$ or $4$. (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: $\sqrt{4} = 2 \neq 2 - 4 = -2$, 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 7

1. $\{6\}$ (correct answer)
2. $\{3\}$
3. $\{-6\}$
4. $\{0\}$

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 $\frac{x}{x-3}=2$: (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 8

1. $x = 2$ only, since this makes the denominator $x-2$ equal to zero
2. $x = -2$ only, since this makes the denominator $x+2$ equal to zero
3. $x = 2$ or $x = -2$, since these values make various denominators equal to zero (correct answer)
4. $x = 4$ 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 $x = 2$ or $x = -2$. 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 $x = 2$ or $x = -2$. 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 9

1. $x^2 + 5x = x^2 + 5$, which simplifies to $5x = 5$ (correct answer)
2. $x^2 - 5x = x^2 + 5$, which simplifies to $-5x = 5$
3. $5x^2 + x = x^2 + 5$, which simplifies to $4x^2 + x - 5 = 0$
4. $3x^2 + x = x^2 + 5$, 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 10

1. Both solutions are valid since they both make the radicand $2x + 3$ positive
2. Only $x = 4 + \sqrt{10}$ is valid since $x = 4 - \sqrt{10}$ makes $x - 3$ negative while the square root is positive (correct answer)
3. Only $x = 4 - \sqrt{10}$ is valid since $x = 4 + \sqrt{10}$ leads to an incorrect equation when substituted back
4. 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 11

1. Check each solution in the original equation since squaring can introduce extraneous solutions (correct answer)
2. Check each solution in the squared equation since the original equation might have been incorrect
3. Accept all solutions since squaring both sides preserves the solution set completely
4. 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 12

1. $x + 12 = 4 + x + 4\sqrt{x}$
2. $x + 12 = 4 + x$
3. $12 = 4 + 4\sqrt{x}$ (correct answer)
4. $12 = 4\sqrt{x}$

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 13

1. $\{-3, 3\}$ with no restrictions needed
2. $\{-3\}$ only, since $x = 3$ makes the original equation undefined (correct answer)
3. $\{3\}$ only, since $x = -3$ is extraneous to the original equation
4. 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 $x = -3$.

Question 14

1. $\{3\}$ (correct answer)
2. $\{1\}$
3. $\{0\}$
4. $\{3,1\}$

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 15

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

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 $\sqrt{x + 3} = 5$), 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 16

1. $\{2\}$ (correct answer)
2. $\{1\}$
3. $\{4\}$
4. $\{2,1\}$

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 $3/(x-1) + 2 = 5$: (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 17

1. $\{0,1\}$ (both valid) (correct answer)
2. $\{1\}$ (and $x=0$ is extraneous)
3. $\{0\}$ (and $x=1$ is extraneous)
4. $\varnothing$

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 $\sqrt{x + 3} = 5$), 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: $x = 0$ or $x = 1$. (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 18

1. $\left\{4\right\}$
2. No real solution
3. $\left\{9\right\}$ (correct answer)
4. $\left\{4,\,9\right\}$

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 $\sqrt{x + 3} = 5$), 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 $\sqrt{x} = x - 6$: (1) The radical is already isolated. (2) Square both sides: $(\sqrt{x})^2 = (x - 6)^2 \rightarrow x = x^2 - 12x + 36$. (3) Rearrange and solve: $0 = x^2 - 13x + 36 = (x - 4)(x - 9) \rightarrow x = 4 \text{ or } x = 9$. (4) Check each in ORIGINAL equation: For x = 4: $\sqrt{4} = 2$, but $x - 6 = 4 - 6 = -2$. Since $2 \neq -2$, this is extraneous! For x = 9: $\sqrt{9} = 3$ and $x - 6 = 9 - 6 = 3$. Since $3 = 3$, this is valid! Final answer: x = 9 only. Choice B correctly identifies that only x = 9 is valid, rejecting x = 4 as extraneous because it doesn't satisfy the original equation. Choice A includes x = 4 as a valid solution, but checking: $\sqrt{4} = 2$ while $4 - 6 = -2$, so $2 \neq -2$. This is extraneous! Always verify: For radical equations, does substituting back work? If it doesn't match, reject the solution. Why extraneous solutions happen: when we square $\sqrt{x} = x - 6$, we get $x = (x - 6)^2$, which is satisfied by both x = 4 and x = 9. But in the ORIGINAL $\sqrt{x} = x - 6$, we need the left side (always non-negative) to equal the right side. When x = 4, we get $\sqrt{4} = 2$ on the left but $-2$ on the right—impossible! The squaring hid this sign issue. These operations are legal but can introduce extras—checking filters them out.

Question 19

1. $x=2$
2. $x=18$ (correct answer)
3. $x=25$
4. 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 $√(...) = \text{negative}$, 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 $√(x + 7) - 2 = 3$: (1) Isolate the radical: $√(x + 7) = 5$. (2) Square both sides: $(√(x + 7))^2 = 5^2 \rightarrow x + 7 = 25$. (3) Solve: $x = 25 - 7 = 18$. (4) Check in ORIGINAL equation: For x = 18: $√(18 + 7) - 2 = √25 - 2 = 5 - 2 = 3$ ✓. This is valid because both sides equal 3. Final answer: x = 18. Choice B correctly solves to get x = 18 and verifies it satisfies the original equation with proper checking. Choice C might result from an arithmetic error when solving $x + 7 = 25$, getting x = 25 instead of x = 18. With the multiple steps in these equations (isolate, square, solve, check), arithmetic errors can compound. Take your time with each step! 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 20

1. $\left\{-\frac{4}{3},\,2\right\}$ (correct answer)
2. $\left\{-\frac{4}{3}\right\}$
3. $\left\{0,\,2\right\}$
4. $\left\{\frac{4}{3},\,-2\right\}$

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 $\frac{1}{x}+\frac{1}{x+2}=\frac{3}{4}$: (1) Identify the LCD: $4x(x+2)$. (2) Multiply every term by the LCD: $4(x+2) + 4x = 3x(x+2)$. (3) Simplify: $4x + 8 + 4x = 3x^2 + 6x$, which gives $8x + 8 = 3x^2 + 6x$. Rearranging: $3x^2 - 2x - 8 = 0$. (4) Factor or use quadratic formula: $(3x + 4)(x - 2) = 0$, so $x = -\frac{4}{3}$ or $x = 2$. (5) Check in original: For $x = -\frac{4}{3}$: denominators are $-\frac{4}{3}$ and $\frac{2}{3}$, neither zero. ✓ For $x = 2$: denominators are $2$ and $4$, neither zero. ✓ Both are valid! Final answer: $\{-\frac{4}{3}, 2\}$. Choice A correctly solves to get both solutions $\{-\frac{4}{3}, 2\}$ and verifies neither makes any denominator zero. Choice B only includes $x = -\frac{4}{3}$ but misses the valid solution $x = 2$. When solving quadratic equations that arise from rational equations, make sure to find ALL solutions and check each one—don't stop at the first solution you find! Quick verification trick: when checking a rational equation solution, list each denominator, substitute the solution, verify none equal zero. For this problem: denominators are $x$ and $x+2$. Neither $-\frac{4}{3}$ nor $2$ makes these zero. Organized checking prevents missing valid solutions or including extraneous ones!