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!

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!

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!

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!

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!

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!

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!

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 $x^2 - 4$.

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.

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.