Explaining Extraneous Solutions - Math 3

When solving $\sqrt{x^2-9} = x-3$ , a student squares both sides to get $x^2-9 = (x-3)^2$ , which simplifies to $x^2-9 = x^2-6x+9$ , leading to $x = 3$ . However, substituting $x = 3$ into the original equation yields $0 = 0$ , which seems to confirm the solution, yet further analysis reveals an issue. What subtle problem explains why $x = 3$ might be considered extraneous?
Answer: The solution $x = 3$ is actually valid, and the appearance of being extraneous comes from misunderstanding that $0 = 0$ represents a true equality confirming the solution works in the original equation. The solution $x = 3$ is actually valid, not extraneous. When $x = 3$ : $\sqrt{3^2-9} = \sqrt{0} = 0$ and $x-3 = 3-3 = 0$ , so $0 = 0$ is true. The student's concern about this being extraneous is unfounded. The domain requires $x^2-9 \geq 0$ , which gives $x \leq -3$ or $x \geq 3$ , and $x = 3$ satisfies this. Choice A correctly identifies the domain but wrongly suggests $x = 3$ is problematic. Choice B incorrectly claims the identity masks indeterminacy. Choice C wrongly suggests the equation is undefined at $x = 3$ .
When solving $\frac{x^2 - 4}{x - 2} = x + 2$ , a student factors the numerator to get $\frac{(x - 2)(x + 2)}{x - 2} = x + 2$ , then cancels the $(x - 2)$ terms to obtain $x + 2 = x + 2$ . They conclude this is an identity true for all real numbers. However, when graphing both sides of the original equation, they notice the graphs don't coincide everywhere. Which analysis best explains why their algebraic conclusion differs from the graphical evidence?
Answer: The cancellation of $(x - 2)$ terms was invalid because it assumed $x - 2 \neq 0$ , effectively removing $x = 2$ from the domain. When you encounter rational equations involving factoring and cancellation, always pay close attention to domain restrictions. The key insight here is understanding what happens to the domain when you perform algebraic operations. The student's algebra is technically correct: $x^2 - 4 = (x-2)(x+2)$ , so $\frac{(x-2)(x+2)}{x-2} = x+2$ when $x \neq 2$ . However, this cancellation is only valid when $x - 2 \neq 0$ , meaning $x \neq 2$ . The original equation $\frac{x^2-4}{x-2} = x+2$ is undefined at $x = 2$ because division by zero is impossible. When you cancel the $(x-2)$ terms, you're implicitly assuming $x \neq 2$ , which removes this point from consideration entirely. The graphs don't coincide everywhere because the left side has a hole at $x = 2$ (since the function is undefined there), while $y = x + 2$ is a continuous line with no breaks. At $x = 2$ , the right side equals 4, but the left side doesn't exist. Answer D correctly identifies that the cancellation assumes $x - 2 \neq 0$ , effectively removing $x = 2$ from the domain. Answer A incorrectly suggests extraneous solutions when the issue is domain restriction. Answer B contains a factoring error— $x^2 - 4$ definitely equals $(x-2)(x+2)$ . Answer C wrongly blames graphing software when the discontinuity is mathematically real. Strategy tip: Before canceling common factors in rational equations, always note what values make those factors zero—these values are excluded from the domain and won't be solutions.
While solving $\sqrt{x + 7} - \sqrt{x - 2} = 3$ , a student isolates one radical: $\sqrt{x + 7} = 3 + \sqrt{x - 2}$ , then squares both sides to get $x + 7 = 9 + 6\sqrt{x - 2} + (x - 2)$ . After simplifying and isolating the remaining radical, they square again and eventually find $x = 2$ . When checked, this value makes $\sqrt{x - 2} = 0$ , but the original equation becomes $3 - 0 = 3$ , which is true. However, $x = 2$ is at the boundary of the domain. What is the most important consideration for determining if extraneous solutions might arise in this type of problem?
Answer: Whether the squaring operations preserve the signs of both sides of the equation at each step of the solution process. The correct answer is A. The key issue with radical equations involving multiple squaring operations is that squaring can change the sign relationships between the two sides of an equation. Each time we square, we must ensure that both sides have the same sign (or at least that we account for sign changes). In this problem, when we have √(x + 7) = 3 + √(x - 2), both sides are non-negative (since we're adding 3 to a square root), so squaring is valid. However, the critical consideration is whether each squaring step preserves the directional equality. If at any point the signs don't match, we can introduce extraneous solutions. B is incorrect because domain considerations alone don't explain extraneous solutions—solutions can be in the domain but still extraneous. C is too general and doesn't specifically address the mechanism by which extraneous solutions arise in radical equations. D is incorrect because the nature of the solutions (integer vs. non-integer) doesn't determine whether they're extraneous.
Consider the equation $x^4 - 13x^2 + 36 = 0$ . A student uses substitution $u = x^2$ to get $u^2 - 13u + 36 = 0$ , which factors as $(u-4)(u-9) = 0$ . This gives $u = 4$ or $u = 9$ , so $x^2 = 4$ or $x^2 = 9$ , yielding $x = \pm 2$ or $x = \pm 3$ . All four solutions check out in the original equation. A peer suggests that two of these solutions should be considered extraneous because the substitution method can introduce false solutions. How should this suggestion be evaluated?
Answer: The peer is incorrect; the substitution $u = x^2$ is a valid one-to-one transformation for $u \geq 0$ , and reversing it by taking $x = \pm\sqrt{u}$ legitimately produces all possible solutions. The peer is incorrect. The substitution $u = x^2$ is valid and doesn't introduce extraneous solutions. When we find $u = 4$ and $u = 9$ , we correctly reverse the substitution: $x^2 = 4$ gives $x = \pm 2$ , and $x^2 = 9$ gives $x = \pm 3$ . All four solutions are legitimate roots of the original fourth-degree polynomial. The fact that we get four solutions is expected for a quartic equation. Choice A incorrectly assumes substitution always risks extraneous solutions. Choice C misunderstands polynomial degree and complex solutions. Choice D correctly notes that forward substitution doesn't introduce extraneous solutions but doesn't fully address the peer's concern.
When solving $\sqrt{2x+3} + x = 6$ , a student rearranges to $\sqrt{2x+3} = 6-x$ and squares both sides: $2x+3 = (6-x)^2 = 36-12x+x^2$ . This yields $x^2-14x+33 = 0$ , which factors as $(x-3)(x-11) = 0$ , giving $x = 3$ or $x = 11$ . Upon checking, $x = 3$ works but $x = 11$ does not. The student concludes that $x = 11$ is extraneous because it makes $6-x = -5 < 0$ . What additional insight explains why this negativity specifically creates the extraneous solution?
Answer: When $6-x = -5$ , the squaring operation converts this to $(-5)^2 = 25$ , but the original equation would require $\sqrt{2x+3} = -5$ , which is impossible since square roots are non-negative. When $x = 11$ , we get $6-x = 6-11 = -5$ . The squared equation becomes $\sqrt{2(11)+3} = \sqrt{25} = 5$ and $(6-x)^2 = (-5)^2 = 25\$, so$ $5^2$ = 25 $checks out in the squared version. However, the original equation requires$ \sqrt{25} = -5 $, or$ 5 = -5 $, which is false. The squaring operation masked the sign difference because$ $(-5)^2$ = $5^2$ $. Choice A is correct in spirit but less precise. Choice B incorrectly suggests a domain restriction on$ x$$. Choice D dramatically mischaracterizes the situation with infinity language.
A student solves $2^{x+1} = 2^{x^2-3}$ by using the property that if $a^m = a^n$ then $m = n$ (for $a > 0, a \neq 1$ ), obtaining $x+1 = x^2-3$ . This gives the quadratic $x^2-x-4 = 0$ with solutions $x = \frac{1 \pm \sqrt{17}}{2}$ . Both solutions check out when substituted back into the original equation. A classmate claims that one solution should be extraneous because exponential equations typically have unique solutions. How should this claim be evaluated?
Answer: The classmate is incorrect; the property $a^m = a^n \Rightarrow m = n$ is valid for all real exponents when $a > 0$ and $a \neq 1$ , so both solutions are legitimate. The classmate is incorrect. Both solutions are valid because the reasoning is sound: if $2^{x+1} = 2^{x^2-3}$ , then the exponents must be equal (since the exponential function with base 2 is one-to-one), giving $x+1 = x^2-3$ . This quadratic can legitimately have two solutions, each representing a value of $x$ where the two exponential expressions are equal. There's no mathematical principle requiring exponential equations to have unique solutions when the exponents involve polynomials. Choice A incorrectly limits exponential equations to one solution. Choice C misapplies monotonicity. Choice D is correct in conclusion but uses imprecise language about 'intersection points of exponential curves.'
A student solves the equation $\sqrt{x + 3} = x - 1$ by squaring both sides to get $x + 3 = (x - 1)^2$ . After expanding and simplifying, they find $x = 1$ or $x = 6$ . When checking their solutions, they discover that $x = 1$ is extraneous. Which statement best explains why this extraneous solution arose?
Answer: Squaring both sides introduced a solution that satisfies the squared equation but not the original equation's domain restrictions. The correct answer is A. Extraneous solutions arise because squaring both sides of an equation is not an equivalent transformation—it can introduce solutions. When x = 1, the right side of the original equation becomes 1 - 1 = 0, but the left side becomes √(1 + 3) = 2. Since 2 ≠ 0, x = 1 doesn't satisfy the original equation. However, when both sides are squared, (2)² = (0)² becomes 4 = 0, which is false, but the algebraic process that led to x = 1 was based on the squared form x + 3 = (x - 1)². The issue isn't domain restrictions of the square root (since x + 3 ≥ 0 when x = 1), but rather that squaring can make a false statement (√4 = 0) appear algebraically valid in the squared form. B is incorrect because squaring doesn't always create extraneous solutions. C is incorrect because the algebra was performed correctly. D is incorrect because the issue isn't about eliminating negative solutions.
A student solves $\sqrt{x+1} - \sqrt{x-3} = 2$ by isolating one radical: $\sqrt{x+1} = 2 + \sqrt{x-3}$ , then squaring both sides to get $x+1 = 4 + 4\sqrt{x-3} + (x-3)$ . This simplifies to $\sqrt{x-3} = 0$ , giving $x = 3$ . However, when $x = 3$ is substituted into the original equation, we get $2 - 0 = 2$ , which is true. Despite this, some argue that $x = 3$ should be considered extraneous. What reasoning supports this view?
Answer: The solution $x = 3$ is actually valid; the argument for it being extraneous stems from a misunderstanding of how boundary values function in radical equations. The solution $x = 3$ is actually valid. When $x = 3$ : $\sqrt{3+1} - \sqrt{3-3} = \sqrt{4} - \sqrt{0} = 2 - 0 = 2$ , which satisfies the original equation. The domain requires $x \geq 3$ (so that $\sqrt{x-3}$ is defined), and $x = 3$ satisfies this. There's no mathematical reason to consider this solution extraneous just because it's a boundary value. Choice A incorrectly suggests boundary conditions are problematic. Choice B mentions constraint-addition but this doesn't make the solution invalid. Choice D mischaracterizes the situation as creating an identity rather than a solution.
Consider solving $\frac{x-1}{x+2} = \frac{x-1}{x-3}$ by multiplying both sides by $(x+2)(x-3)$ to get $(x-1)(x-3) = (x-1)(x+2)$ . A student then divides both sides by $(x-1)$ to obtain $x-3 = x+2$ , which gives $-3 = 2$ , suggesting no solution exists. However, direct substitution shows that $x = 1$ satisfies the original equation. What explains this apparent contradiction?
Answer: The division by $(x-1)$ was invalid because this expression equals zero when $x = 1$ , and dividing by zero eliminates valid solutions from consideration during the algebraic process. When $x = 1$ , we get $(x-1)(x-3) = (x-1)(x+2)$ becoming $0 \cdot (-2) = 0 \cdot 3$ , or $0 = 0$ , which is true. The student's error was dividing both sides by $(x-1)$ when $(x-1) = 0$ . This division is invalid and eliminates the solution $x = 1$ . In the original equation, when $x = 1$ : $\frac{1-1}{1+2} = \frac{1-1}{1-3}$ gives $\frac{0}{3} = \frac{0}{-2}$ , or $0 = 0$ , which is true. Choice B incorrectly describes the mechanism. Choice C wrongly claims the equation is undefined. Choice D incorrectly accepts the invalid algebraic step.
Consider the equation $\log(x - 1) + \log(x + 2) = \log(8)$ . A student uses the logarithm property $\log(a) + \log(b) = \log(ab)$ to rewrite this as $\log[(x - 1)(x + 2)] = \log(8)$ . They then conclude that $(x - 1)(x + 2) = 8$ , leading to $x^2 + x - 2 = 8$ , or $x^2 + x - 10 = 0$ . Using the quadratic formula, they find $x = \frac{-1 \pm \sqrt{41}}{2}$ . When checking these solutions, they discover that $x = \frac{-1 - \sqrt{41}}{2} \approx -3.7$ creates undefined logarithms in the original equation. What is the fundamental reason this extraneous solution arose?
Answer: The logarithm property $\log(a) + \log(b) = \log(ab)$ is only valid when both $a > 0$ and $b > 0$ , but the algebraic solution process ignored these domain restrictions. When solving logarithmic equations, you must always consider the domain restrictions of logarithms alongside the algebraic manipulations. Logarithms are only defined for positive arguments, which creates constraints that purely algebraic approaches can miss. The student's algebraic work is mathematically sound: applying $\log(a) + \log(b) = \log(ab)$ , converting to $(x-1)(x+2) = 8$ , and solving the quadratic correctly. However, the logarithm property $\log(a) + \log(b) = \log(ab)$ requires both $a > 0$ and $b > 0$ . This means we need $x - 1 > 0$ and $x + 2 > 0$ , so $x > 1$ . The solution $x = \frac{-1 - \sqrt{41}}{2} \approx -3.7$ makes $x - 1 \approx -4.7 < 0$ , violating the domain restriction. When this value is substituted back, $\log(x-1)$ becomes $\log(\text{negative number})$ , which is undefined in the real number system. Choice A is wrong because the algebra was performed correctly. Choice B incorrectly suggests the conversion from $\log[(x-1)(x+2)] = \log(8)$ to $(x-1)(x+2) = 8$ is problematic—this step is valid when both sides are defined. Choice C is incorrect because both solutions are real numbers; the issue isn't about complex components. Choice D correctly identifies that ignoring domain restrictions when applying logarithm properties leads to extraneous solutions. Study tip: Always check domain restrictions first in logarithmic equations. The valid domain often eliminates solutions before you even solve algebraically.
A student solves $\sqrt[3]{x + 1} = -2$ by cubing both sides to obtain $(x + 1) = (-2)^3 = -8$ , leading to $x = -9$ . When they check this solution by substituting back into the original equation, they get $\sqrt[3]{-9 + 1} = \sqrt[3]{-8} = -2$ , which matches the right side. However, they've heard that radical equations often produce extraneous solutions and wonder if they should reject $x = -9$ . What is the best explanation regarding extraneous solutions in this context?
Answer: No extraneous solution exists because cubing both sides of an equation is a reversible operation that preserves all solutions. The correct answer is B. Unlike squaring, which can introduce extraneous solutions because it's not a one-to-one operation (both positive and negative numbers square to positive results), cubing is a one-to-one operation that preserves all solutions. When we cube both sides of an equation, we create an equivalent equation—no solutions are gained or lost. In this case, x = -9 is a valid solution because cube roots of negative numbers are well-defined in the real number system (∛(-8) = -2). The student's verification confirms this. A is incorrect because cube roots of negative numbers are defined (∛(-8) = -2). C is incorrect because cube roots don't have domain restrictions like square roots—they're defined for all real numbers. D is incorrect because we're working in the real number system, and the cube root function is single-valued for real inputs. The key insight is that extraneous solutions typically arise from operations like squaring that are not one-to-one, but cubing preserves the solution set.
When solving $\sqrt[3]{x+5} = \sqrt[3]{2x-1}$ , a student cubes both sides to get $x+5 = 2x-1$ , which gives $x = 6$ . This solution checks out perfectly in the original equation: $\sqrt[3]{6+5} = \sqrt[3]{11}$ and $\sqrt[3]{2(6)-1} = \sqrt[3]{11}$ . However, the student worries that cubing both sides might have introduced an extraneous solution, similar to what happens when squaring both sides of equations. Is this concern justified?
Answer: No, the concern is not justified because the cube root function is defined for all real numbers and the cubing operation preserves the equality relationship without introducing extraneous solutions. When working with radical equations, you need to understand which operations preserve equality and which might introduce false solutions. The key insight here is recognizing the difference between even and odd roots. The correct answer is D because cubing both sides of an equation is a completely reversible operation that preserves equality. Unlike squaring, which can create extraneous solutions, cubing maintains a one-to-one relationship. Since the cube root function $\sqrt[3]{x}$ is defined for all real numbers (positive, negative, and zero) and is strictly increasing, if $\sqrt[3]{a} = \sqrt[3]{b}$ , then $a = b$ must be true. When you cube both sides, you're simply undoing the cube root operation without any ambiguity. Answer A is incorrect because it falsely equates cubing with squaring. While squaring can introduce extraneous solutions (since both positive and negative numbers have the same square), cubing preserves the sign and magnitude relationship. Answer B correctly identifies that cube roots handle negative arguments, but it misses the fundamental point. The issue isn't just about negative arguments—it's about the one-to-one nature of the cubic function itself. Answer C makes the common mistake of assuming all power operations behave the same way. This overlooks the crucial distinction between even and odd powers: even powers can map different inputs to the same output, while odd powers maintain a unique relationship. Study tip: Remember that odd-powered operations (cubing, fifth powers, etc.) never introduce extraneous solutions because they preserve the one-to-one relationship between inputs and outputs. Only even-powered operations require checking for extraneous solutions.
A student attempts to solve $\sqrt{x^2} = x - 6$ by recognizing that $\sqrt{x^2} = |x|$ . They then consider two cases: Case 1: If $x \geq 0$ , then $|x| = x$ , so $x = x - 6$ , which gives $0 = -6$ (impossible). Case 2: If $x < 0$ , then $|x| = -x$ , so $-x = x - 6$ , which gives $x = 3$ . However, $x = 3$ contradicts the assumption $x < 0$ for Case 2. The student concludes there are no solutions. Which statement best explains the error in reasoning that led to this incorrect conclusion?
Answer: Case 2 analysis was performed correctly, but finding $x = 3$ while assuming $x < 0$ indicates this case doesn't apply, and the student should recognize that Case 1 already showed no solutions exist. The correct answer is C. The student's approach using |x| = √(x²) is correct, and their case analysis is properly executed. The key insight is in interpreting what it means when a case yields a solution that violates its assumption. When Case 2 (assuming x < 0) produces x = 3, this doesn't mean x = 3 is a solution—it means Case 2 doesn't apply to any actual solutions. Since Case 1 already showed no solutions exist (0 = -6 is impossible), and Case 2 produces a contradiction, the conclusion of no solutions is actually correct. The error is not in the mathematics but in the logical interpretation: the student should recognize that getting x = 3 from Case 2 simply confirms that no values of x < 0 satisfy the equation. A is incorrect because √(x²) = |x| is correct. B is incorrect because testing x = 3 in Case 1 gives 3 = 3 - 6 or 3 = -3, which is false. D would introduce extraneous solutions.
Consider the equation $\log_2(x+4) + \log_2(x-2) = 3$ . A student uses the property $\log_a(m) + \log_a(n) = \log_a(mn)$ to rewrite this as $\log_2[(x+4)(x-2)] = 3$ , then converts to exponential form: $(x+4)(x-2) = 8$ . This yields the quadratic $x^2+2x-16 = 0$ with solutions $x = -1 \pm \sqrt{17}$ . One of these solutions is extraneous. What principle explains why?
Answer: The logarithm property $\log_a(m) + \log_a(n) = \log_a(mn)$ is only valid when both $m > 0$ and $n > 0$ , but one solution makes either $x+4$ or $x-2$ negative. For $\log_2(x+4) + \log_2(x-2) = 3$ to be defined, we need both $x+4 > 0$ and $x-2 > 0$ , which means $x > 2$ . The solutions are $x = -1 + \sqrt{17} \approx 3.12$ and $x = -1 - \sqrt{17} \approx -5.12$ . Since $\sqrt{17} > 4$ , the negative solution gives $x < -5$ , making both $x+4 < 0$ and $x-2 < 0$ , so the logarithms are undefined. Choice B is incorrect because the conversion was done correctly. Choice C is true but doesn't explain the mechanism. Choice D misunderstands how the extraneous solution arose.
A student solves $|2x-5| = 3x-8$ by considering two cases: when $2x-5 \geq 0$ and when $2x-5 < 0$ . From the first case, they get $x = 3$ , and from the second case, they get $x = -3$ . When checking, $x = -3$ doesn't satisfy the original equation. Which analysis best explains why this extraneous solution occurred?
Answer: The solution $x = -3$ makes $3x-8$ negative, but absolute value expressions can only equal non-negative quantities, creating a mathematical impossibility in the original equation. Since $|2x-5| \geq 0$ for all real $x$ , we need $3x-8 \geq 0$ for any valid solution. When $x = -3$ , we get $3(-3)-8 = -17 < 0$ , which means we're trying to solve $|2x-5| = -17$ . This is impossible since absolute values are never negative. Choice A is incorrect because the issue isn't about both cases yielding positive results. Choice B is wrong because $x = -3$ does satisfy $2x-5 < 0$ . Choice D is incorrect because the case work was done properly.
Consider the equation $\frac{x}{x-2} + \frac{3}{x+1} = \frac{x+7}{(x-2)(x+1)}$ . A student multiplies both sides by $(x-2)(x+1)$ to eliminate fractions, then solves the resulting linear equation to find $x = 2$ . Upon substitution back into the original equation, this solution is undefined. What fundamental principle explains why this extraneous solution emerged?
Answer: The clearing of fractions implicitly assumed that $(x-2)(x+1) \neq 0$ , but the resulting solution violates this assumption by making the clearing factor equal to zero. When solving rational equations by clearing fractions, you're performing a conditional algebraic operation that requires careful attention to domain restrictions. The fundamental issue here is that multiplying both sides by $(x-2)(x+1)$ is only valid when this expression is non-zero. This multiplication implicitly assumes that $x \neq 2$ and $x \neq -1$ . However, the algebraic process treats this as an unconditional step, potentially generating solutions that violate the very assumption that made the clearing operation legitimate. When you clear fractions and solve to get $x = 2$ , you've found a solution to the transformed linear equation. But $x = 2$ makes $(x-2)(x+1) = 0$ , which means the clearing step was invalid for this value. The multiplication introduced $x = 2$ as an apparent solution, but it contradicts the condition required for the multiplication to be mathematically sound. Answer choice A incorrectly focuses on polynomial degree changes, but rational equations don't gain roots simply from degree changes. Choice B mentions indeterminate forms, but the issue isn't about numerators becoming infinite—it's about denominators becoming zero. Choice C discusses domain expansion, which is partially correct but misses the specific logical flaw in the clearing process. Choice D correctly identifies that the clearing operation assumed $(x-2)(x+1) \neq 0$ , and the resulting solution directly violates this assumption. Study tip: Always check potential solutions in the original equation's denominators before substituting into the full equation. If any denominator becomes zero, that solution is automatically extraneous.
A student solves the equation $\sqrt{x+3} = x-1$ by squaring both sides to get $x+3 = (x-1)^2$ , which simplifies to $x^2-3x-2=0$ . The solutions are $x=\frac{3\pm\sqrt{17}}{2}$ . When checking these solutions in the original equation, one proves to be extraneous. Which statement best explains why this extraneous solution arose?
Answer: The squaring operation is not reversible and can introduce solutions that satisfy the squared equation but violate the non-negative requirement for the right side of the original equation. When we square both sides of $\sqrt{x+3} = x-1$ , we lose the information that $x-1$ must be non-negative (since square roots yield non-negative values). The squaring creates an equation that's satisfied by both $x-1 \geq 0$ and $x-1 < 0$ , but only the former satisfies the original equation. Choice A is incorrect because both solutions keep $x+3 \geq 0$ . Choice C is wrong because the algebra is correct. Choice D is incorrect because we're dealing with real solutions in both equations.
While solving $\sqrt{2x + 3} + x = 6$ , a student rearranges to get $\sqrt{2x + 3} = 6 - x$ , then squares both sides to obtain $2x + 3 = (6 - x)^2 = 36 - 12x + x^2$ . This simplifies to $x^2 - 14x + 33 = 0$ , which factors as $(x - 3)(x - 11) = 0$ , giving $x = 3$ and $x = 11$ . Checking these solutions: when $x = 3$ , the original equation becomes $\sqrt{9} + 3 = 3 + 3 = 6$ ✓; when $x = 11$ , it becomes $\sqrt{25} + 11 = 5 + 11 = 16 \neq 6$ ✗. The student recognizes that $x = 11$ is extraneous. Which analysis most precisely explains why $x = 11$ became an extraneous solution?
Answer: The value $x = 11$ makes $6 - x = -5 < 0$ , but $\sqrt{2x + 3} = 5 > 0$ , violating the equality $\sqrt{2x + 3} = 6 - x$ that was squared. The correct answer is B. The crucial insight is that when we squared both sides of √(2x + 3) = 6 - x, we were assuming that both sides had the same sign. Square roots are always non-negative, so √(2x + 3) ≥ 0. For the equation to be valid, we also need 6 - x ≥ 0, which means x ≤ 6. When x = 11, we have 6 - x = 6 - 11 = -5, which is negative. However, √(2x + 3) = √(22 + 3) = √25 = 5, which is positive. Since 5 ≠ -5, the equation √(2x + 3) = 6 - x is not satisfied when x = 11. The squaring operation masked this sign inconsistency because (-5)² = 5² = 25. This is why x = 11 satisfies the squared equation but not the original. A is incorrect because it doesn't clearly explain the sign issue. C is incorrect because the domain restriction 2x + 3 ≥ 0 (i.e., x ≥ -3/2) is satisfied by both solutions. D is incorrect because factoring doesn't introduce new roots.

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