Solving Rational and Radical Equations - Algebra 1

Solve and check for extraneous solutions: $\frac{x}{x-3}=2.$
Answer: $\{6\}$ 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 $x/(x-3) = 2$ : (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!
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)?
Answer: $\{7\}$ only 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! Testing the proposed solutions in the original equation $\sqrt{x+2}=x-4$ : 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!
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?
Answer: $\{-3\}$ only, since $x = 3$ makes the original equation undefined 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 $x = -3$ .
What is the solution set (excluding extraneous) for $\frac{x+1}{x-1}=2?$
Answer: $\{3\}$ 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!
Solve and check for extraneous solutions: $\sqrt{x+3}=x-3.$
Answer: $\{6\}$ only (and $x=1$ is extraneous) 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!
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:
Answer: $x = 2$ or $x = -2$ , since these values make various denominators equal to zero 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 $x^2 - 4$ .
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?
Answer: Only $x = 4 + \sqrt{10}$ is valid since $x = 4 - \sqrt{10}$ makes $x - 3$ negative while the square root is positive 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.
Solve and check for extraneous solutions: $\frac{3}{x-1}+2=5.$
Answer: $\{2\}$ 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!
Solve and check for extraneous solutions: $\sqrt{x+5}+1=x.$
Answer: $\{4\}$ 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!
Solve and check for extraneous solutions: $\frac{x}{x-3}=2$
Answer: $\{6\}$ 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!
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?
Answer: Check each solution in the original equation since squaring can introduce extraneous solutions 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.
Solve and verify your answer: $\sqrt{3x+1}=x+1.$
Answer: $\{0,1\}$ (both valid) 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.
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)?
Answer: Only $x=6$ is valid; $x=1$ is extraneous 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!
Solve and check for extraneous solutions: $\frac{1}{x}+\frac{1}{x+2}=\frac{3}{4}$
Answer: $\left\{-\frac{4}{3},\,2\right\}$ 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!
Solve and check for extraneous solutions: $\frac{3}{x-2}=2$
Answer: $x=\frac{7}{2}$ 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!
Solve and check for extraneous solutions: $\sqrt{x} = 2 - x.$
Answer: $\{1\}$ only (and $x=4$ is extraneous) 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!
Solve and check for extraneous solutions: $\sqrt{x}=x-6$
Answer: $\left\{9\right\}$ 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.
Solve and check for extraneous solutions: $\frac{x}{x-2}=3$
Answer: $x=3$ 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!
Solve and check for extraneous solutions: $\sqrt{x+5}+1=x.$
Answer: $\{4\}$ 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! Solving $\sqrt{x+5} + 1 = x$ : (1) Isolate radical: $\sqrt{x+5} = x - 1$ . (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 $: $\sqrt{4+5} + 1 = \sqrt{9} + 1 = 3 + 1 = 4$ , matches $ x=4 $, valid. For $ x=-1 $: $\sqrt{-1+5} + 1 = \sqrt{4} + 1 = 2 + 1 = 3 \neq x=-1$ , 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!
Solve and check for extraneous solutions: $\frac{x}{x-5}=3$
Answer: $x=\frac{15}{2}$ 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!

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