Q: Compare greater roots (definition repaired): Let x be the greater root of x^2 + 10x + 24 = 0 and y be the greater root of 4y^2 − 17y + 18 = 0. Choose the correct relation.

Introduction / Context: We repair the stem by explicitly defining the comparison target: the larger roots of each quadratic. Then we compute those roots and compare. If x is strictly less than y, the statement x ≤ y is true as well. Given Data / Assumptions: x^2 + 10x + 24 = 0; let x be the greater root. 4y^2 − 17y + 18 = 0; let y be the greater root. Concept / Approach: Factor or use the quadratic formula to find each equation’s larger root. Compare them exactly to avoid sign mistakes, then choose the most accurate relation provided among the options. Step-by-Step Solution: x^2 + 10x + 24 = 0 ⇒ (x + 6)(x + 4) = 0 ⇒ roots −6, −4 ⇒ greater root x = −4 4y^2 − 17y + 18 = 0 ⇒ Δ = 289 − 288 = 1 ⇒ roots (17 ± 1)/8 ⇒ y = 2 or 9/4 = 2.25 ⇒ greater root y = 2.25 Comparison: −4 y contradict the numerical comparison; “cannot be established” is incorrect because both larger roots are uniquely determined. Common Pitfalls: Forgetting that the “greater root” for a quadratic with both negative roots is the one closer to zero. Final Answer: x ≤ y

Q: Compare roots from two quadratics (check all combinations): I) 24x^2 + 38x + 15 = 0 II) 12y^2 + 28y + 15 = 0 Decide the correct relationship that always holds between any root x of I and any root y of II.

Introduction / Context: When comparing two quadratics, each with two real roots, we must confirm that a proposed inequality holds across all four pairings (each x with each y). A single counterexample invalidates a universal statement like x ≥ y. Here we solve both quadratics and compare the root sets carefully. Given Data / Assumptions: I) 24x^2 + 38x + 15 = 0 II) 12y^2 + 28y + 15 = 0 All comparisons are between any root x of I and any root y of II. Concept / Approach: Compute exact roots via discriminants. Then order the roots numerically and check whether the smallest x is still ≥ the largest y (for x ≥ y), or similar tests for other relations. If equality can occur for some pairing and strict inequality for others, then ≥ (not >) may be the correct consistent relation. Step-by-Step Solution: I) D = 38^2 − 4*24*15 = 1444 − 1440 = 4 ⇒ √D = 2x = (−38 ± 2)/(48) ⇒ x ∈ {−36/48, −40/48} = {−3/4, −5/6}II) D = 28^2 − 4*12*15 = 784 − 720 = 64 ⇒ √D = 8y = (−28 ± 8)/(24) ⇒ y ∈ {−20/24, −36/24} = {−5/6, −3/2}Order: x-roots = {−0.75, −0.833…}; y-roots = {−0.833…, −1.5}Check: smallest x = −0.833… ≥ largest y = −0.833… (true, equality), and any x ≥ any y (true) Verification / Alternative check: Test all pairings: (−0.75 ≥ −0.833…), (−0.75 ≥ −1.5), (−0.833… ≥ −0.833…), (−0.833… ≥ −1.5). All hold; at least one pairing yields equality, so x ≥ y is the strongest always-true statement. Why Other Options Are Wrong: x > y: Fails at equality case (−5/6 vs −5/6). x ≤ y or x and ≥. Final Answer: x ≥ y

Question 1

Express α^2 + β^2 in terms of a, b, c: If α and β are roots of ax^2 + bx + c = 0 (a ≠ 0), find α^2 + β^2.

Accepted Answer

Introduction / Context: This is a direct application of Vieta’s relations and algebraic identities. The sum and product of roots of a quadratic allow you to express symmetric functions like α^2 + β^2 without solving for the roots explicitly. Given Data / Assumptions: α and β are roots of ax^2 + bx + c = 0. a ≠ 0 so that degree and relations are valid. Concept / Approach: Use the identity α^2 + β^2 = (α + β)^2 − 2αβ. From Vieta, α + β = −b/a and αβ = c/a. Substitute and simplify carefully to avoid sign or denominator errors. Step-by-Step Solution: α + β = −b/a ⇒ (α + β)^2 = b^2/a^2 αβ = c/a Therefore α^2 + β^2 = (α + β)^2 − 2αβ = (b^2/a^2) − 2(c/a) Put over a common denominator a^2: α^2 + β^2 = (b^2 − 2ac)/a^2 Verification / Alternative check: Pick a sample quadratic (e.g., x^2 − 5x + 6 = 0 with roots 2 and 3). Compute α^2 + β^2 = 4 + 9 = 13. Formula gives (25 − 12)/1 = 13. Checks out. Why Other Options Are Wrong: The ones with +2ac use the wrong sign; others have incorrect scaling or extraneous factors in the denominator. Common Pitfalls: Missing the minus sign in the identity and forgetting to square the −b/a properly. Final Answer: (b^2 − 2ac) / a^2

Question 2

Find an equivalent equation to x^2 − 6x + 5 = 0 (same solution set): Choose the equation that has exactly the same set of real solutions as x^2 − 6x + 5 = 0.

Accepted Answer

Introduction / Context: Two equations are equivalent (have the same solution set) if each solution to one satisfies the other and vice versa. The quadratic x^2 − 6x + 5 = 0 factors neatly, so we can convert it to an absolute-value statement describing the same two points on the number line. Given Data / Assumptions: Original: x^2 − 6x + 5 = 0 We need an equation with exactly the same real roots. Concept / Approach: Factor first: x^2 − 6x + 5 = (x − 1)(x − 5) = 0 ⇒ x = 1 or x = 5. The midpoint of the roots is 3 and each root lies 2 units from 3. That fact is captured by |x − 3| = 2. Step-by-Step Solution: x^2 − 6x + 5 = 0 ⇒ (x − 1)(x − 5) = 0 ⇒ x = 1 or x = 5 Distance form: both solutions are exactly 2 away from 3 ⇒ |x − 3| = 2 Verification / Alternative check: Plug x = 1 and x = 5 into |x − 3| = 2: both satisfy it. The other polynomial options have different root sets (e.g., x^2 − 5x + 6 has roots 2 and 3). Why Other Options Are Wrong: They yield different roots: coefficient changes alter the solution set. Only the absolute-value equation encodes exactly the same two points. Common Pitfalls: Assuming multiplying by a constant keeps roots when constants shift the equation incorrectly; equivalence requires identical solution sets. Final Answer: |x − 3| = 2

Question 3

Compare greater roots from two quadratics (definition repaired): Let x be the greater root of 6x^2 + 41x + 63 = 0 and let y be the greater root of 4y^2 + 8y + 3 = 0. Choose the correct relation between x and y.

Accepted Answer

Introduction / Context: The original “I/II” style stem lacked the comparison rule. Using the Recovery-First Policy, we define x and y as the greater roots of the two quadratics and compare them. This is a common exam pattern in quantity comparison questions. Given Data / Assumptions: x is the larger root of 6x^2 + 41x + 63 = 0. y is the larger root of 4y^2 + 8y + 3 = 0. Concept / Approach: Compute roots using the quadratic formula. For ax^2 + bx + c = 0, roots are (−b ± √(b^2 − 4ac)) / (2a). Identify the greater root for each equation, then compare numerically (exact fractions avoid rounding errors). Step-by-Step Solution: For 6x^2 + 41x + 63 = 0: Δ = 41^2 − 4*6*63 = 1681 − 1512 = 169 ⇒ √Δ = 13 Roots: (−41 ± 13)/12 ⇒ x1 = (−28)/12 = −7/3, x2 = (−54)/12 = −9/2. Greater root x = −7/3 (≈ −2.333) For 4y^2 + 8y + 3 = 0: Δ = 64 − 48 = 16 ⇒ √Δ = 4 Roots: (−8 ± 4)/8 ⇒ y = −1/2 or y = −3/4. Greater root y = −1/2 (−0.5) Comparison: −7/3 < −1/2 ⇒ x < y Verification / Alternative check: Because both larger roots are negative, the less negative (closer to zero) is the greater number; −0.5 is greater than −2.333… Why Other Options Are Wrong: x ≥ y, x > y, equality, or indeterminate are contradicted by the exact computed values. Common Pitfalls: Picking the more negative value as “larger” by mistake; always compare on the number line carefully. Final Answer: x < y

Question 4

Compare greater roots (definition repaired): Let x be the greater root of x^2 + 10x + 24 = 0 and y be the greater root of 4y^2 − 17y + 18 = 0. Choose the correct relation.

Accepted Answer

Introduction / Context: We repair the stem by explicitly defining the comparison target: the larger roots of each quadratic. Then we compute those roots and compare. If x is strictly less than y, the statement x ≤ y is true as well. Given Data / Assumptions: x^2 + 10x + 24 = 0; let x be the greater root. 4y^2 − 17y + 18 = 0; let y be the greater root. Concept / Approach: Factor or use the quadratic formula to find each equation’s larger root. Compare them exactly to avoid sign mistakes, then choose the most accurate relation provided among the options. Step-by-Step Solution: x^2 + 10x + 24 = 0 ⇒ (x + 6)(x + 4) = 0 ⇒ roots −6, −4 ⇒ greater root x = −4 4y^2 − 17y + 18 = 0 ⇒ Δ = 289 − 288 = 1 ⇒ roots (17 ± 1)/8 ⇒ y = 2 or 9/4 = 2.25 ⇒ greater root y = 2.25 Comparison: −4 < 2.25 ⇒ x < y. Among given options, x ≤ y is true. Verification / Alternative check: Substitute the computed roots back into the original equations to confirm correctness before comparing. Why Other Options Are Wrong: x ≥ y and x > y contradict the numerical comparison; “cannot be established” is incorrect because both larger roots are uniquely determined. Common Pitfalls: Forgetting that the “greater root” for a quadratic with both negative roots is the one closer to zero. Final Answer: x ≤ y

Question 5

Compare greater roots (definition repaired): Let x be the greater root of x^2 − 20x + 91 = 0 and y be the greater root of y^2 − 32y + 247 = 0. Choose the correct relation between x and y.

Accepted Answer

Introduction / Context: To resolve ambiguity, we define x, y as the greater roots of their respective quadratics. We then factor or use the quadratic formula to find the larger root for each and compare the results directly. Given Data / Assumptions: x satisfies x^2 − 20x + 91 = 0 (greater root). y satisfies y^2 − 32y + 247 = 0 (greater root). Concept / Approach: Factorization by integers helps: look for pairs with the correct sum and product. Once each pair of roots is found, pick the larger one and compare numerically. Step-by-Step Solution: x^2 − 20x + 91 = 0 ⇒ (x − 7)(x − 13) = 0 ⇒ roots 7, 13 ⇒ greater root x = 13 y^2 − 32y + 247 = 0 ⇒ (y − 13)(y − 19) = 0 ⇒ roots 13, 19 ⇒ greater root y = 19 Hence 13 < 19 ⇒ x < y ⇒ x ≤ y Verification / Alternative check: Cross-check products: 7*13 = 91 and 13*19 = 247; sums 20 and 32, confirming correct factorization. Why Other Options Are Wrong: x > y and x ≥ y contradict the actual numbers; “cannot be established” is false because both larger roots are uniquely known; equality is false since 13 ≠ 19. Common Pitfalls: Swapping smaller and larger roots; forgetting that both equations have positive distinct roots here. Final Answer: x ≤ y

Question 6

Compare roots from two quadratics (set-based comparison, all cases): I) 3x^2 − 20x − 32 = 0 II) 2y^2 − 3y − 20 = 0 Based on all possible real roots of each equation, determine the correct relationship between x and y.

Accepted Answer

Introduction / Context: This is a quantitative comparison question involving two separate quadratic equations. Each quadratic has two real roots; therefore x can take either of two values from the first equation, and y can take either of two values from the second. The task is to compare every possible x with every possible y and decide whether a single, always-true relationship (such as x < y or x ≥ y) can be asserted. Given Data / Assumptions: I) 3x^2 − 20x − 32 = 0 II) 2y^2 − 3y − 20 = 0 We must consider all real roots, not just one root from each equation. Concept / Approach: Solve both quadratics exactly using the quadratic formula. Then list the root sets for x and y. A determinate comparison (like x > y) is valid only if every possible x is greater than every possible y. If there is any cross-over (some x bigger than some y, but also some x smaller than some y), then no fixed relationship can be established. Step-by-Step Solution: I) D = 20^2 + 4*3*32 = 400 + 384 = 784 ⇒ √D = 28x = (20 ± 28)/(2*3) ⇒ x ∈ {8, −4/3}II) D = (−3)^2 + 4*2*20 = 9 + 80 = 89? (Check sign) Actually: 2y^2 − 3y − 20 = 0 ⇒ D = (−3)^2 − 4*2*(−20) = 9 + 160 = 169 ⇒ √D = 13y = (3 ± 13)/(2*2) ⇒ y ∈ {4, −5/2}Compare sets: x-set {8, −1.333…}; y-set {4, −2.5}Observe: 8 > −2.5 (x can be greater), but −1.333… < 4 (x can be smaller) Verification / Alternative check: To assert x ≥ y, the smallest x (−1.333…) must be ≥ largest y (4), which is false. To assert x ≤ y, the largest x (8) must be ≤ smallest y (−2.5), which is false. Hence, no single relation holds for all pairings. Why Other Options Are Wrong: x < y / x ≤ y: Fails because x = 8 and y = −2.5 give x > y. x > y / x ≥ y: Fails because x = −1.333… and y = 4 give x < y. Common Pitfalls: Comparing only one root from each equation or comparing averages instead of all combinations. In such “both roots” comparisons, check extreme pairings (min x vs max y, max x vs min y) to quickly decide. Final Answer: Relationship between x and y cannot be established

Question 7

Compare roots from two quadratics (check all combinations): I) 24x^2 + 38x + 15 = 0 II) 12y^2 + 28y + 15 = 0 Decide the correct relationship that always holds between any root x of I and any root y of II.

Accepted Answer

Introduction / Context: When comparing two quadratics, each with two real roots, we must confirm that a proposed inequality holds across all four pairings (each x with each y). A single counterexample invalidates a universal statement like x ≥ y. Here we solve both quadratics and compare the root sets carefully. Given Data / Assumptions: I) 24x^2 + 38x + 15 = 0 II) 12y^2 + 28y + 15 = 0 All comparisons are between any root x of I and any root y of II. Concept / Approach: Compute exact roots via discriminants. Then order the roots numerically and check whether the smallest x is still ≥ the largest y (for x ≥ y), or similar tests for other relations. If equality can occur for some pairing and strict inequality for others, then ≥ (not >) may be the correct consistent relation. Step-by-Step Solution: I) D = 38^2 − 4*24*15 = 1444 − 1440 = 4 ⇒ √D = 2x = (−38 ± 2)/(48) ⇒ x ∈ {−36/48, −40/48} = {−3/4, −5/6}II) D = 28^2 − 4*12*15 = 784 − 720 = 64 ⇒ √D = 8y = (−28 ± 8)/(24) ⇒ y ∈ {−20/24, −36/24} = {−5/6, −3/2}Order: x-roots = {−0.75, −0.833…}; y-roots = {−0.833…, −1.5}Check: smallest x = −0.833… ≥ largest y = −0.833… (true, equality), and any x ≥ any y (true) Verification / Alternative check: Test all pairings: (−0.75 ≥ −0.833…), (−0.75 ≥ −1.5), (−0.833… ≥ −0.833…), (−0.833… ≥ −1.5). All hold; at least one pairing yields equality, so x ≥ y is the strongest always-true statement. Why Other Options Are Wrong: x > y: Fails at equality case (−5/6 vs −5/6). x ≤ y or x < y: Contradicted by −0.75 ≥ −0.833… Relationship cannot be determined: We just established a universal relation. Common Pitfalls: Overlooking equal roots across equations or assuming strict inequality without testing equality. Always check extremes to decide between > and ≥. Final Answer: x ≥ y

Question 8

Parameter k from a root-sum constraint: For x^2 − 6x + k = 0 with roots α and β, given 3α + 2β = 20, find the value of k.

Accepted Answer

Introduction / Context: This problem uses elementary symmetric relationships of quadratic roots: for x^2 − 6x + k = 0, the sum of roots is α + β = 6 and the product is αβ = k. A linear constraint involving the roots (3α + 2β = 20) lets us determine the individual roots and hence the parameter k. Given Data / Assumptions: Equation: x^2 − 6x + k = 0 Roots: α, β Condition: 3α + 2β = 20 Sum α + β = 6; product αβ = k Concept / Approach: Express β from the sum (β = 6 − α), substitute into 3α + 2β = 20, and solve for α. Then get β and compute k = αβ. This is a direct application of Vieta’s formulas combined with a linear relation. Step-by-Step Solution: β = 6 − α3α + 2(6 − α) = 20 ⇒ 3α + 12 − 2α = 20 ⇒ α = 8β = 6 − 8 = −2k = αβ = 8 * (−2) = −16 Verification / Alternative check: Check the given condition: 3*8 + 2*(−2) = 24 − 4 = 20 (satisfied). Check sum: 8 + (−2) = 6 (matches Vieta). Everything is consistent. Why Other Options Are Wrong: 8, −8, 16, 0: None equal αβ with (α, β) = (8, −2). Common Pitfalls: Forgetting that α + β is fixed by the coefficient of x. Do not attempt to solve via quadratic formula first; the linear relation makes it faster. Final Answer: −16

Question 9

Roots in a given ratio: Find the positive value of m such that the roots of 12x^2 + m x + 5 = 0 are in the ratio 3 : 2.

Accepted Answer

Introduction / Context: When roots are in a known ratio, we can parametrize them as 3t and 2t. Using Vieta’s formulas for a quadratic ax^2 + bx + c = 0 (sum = −b/a, product = c/a), we can solve for t and then deduce the parameter m. The “positive value” clause guides the sign choice for t and hence m. Given Data / Assumptions: Equation: 12x^2 + m x + 5 = 0 Roots in ratio 3 : 2 ⇒ roots 3t and 2t Sum = −m/12; Product = 5/12 Concept / Approach: Set up: 3t + 2t = 5t = −m/12 and (3t)(2t) = 6t^2 = 5/12. First get t^2 from the product, then use the sum to find m. Finally select the sign of t that yields the required positive m (because m depends linearly on t with a negative coefficient in the sum relation). Step-by-Step Solution: 6t^2 = 5/12 ⇒ t^2 = 5/725t = −m/12 ⇒ m = −60tChoose t < 0 to make m > 0 (since m = −60t)t = −√(5/72) = −(√5)/(6√2)m = −60t = 60*(√5)/(6√2) = 10 * √(5/2) = 5√10 Verification / Alternative check: With m = 5√10, sum = −m/12 < 0, matching 5t with t negative; product remains 5/12 > 0. The roots keep the 3:2 ratio in magnitude and are both negative, which is permissible; only m’s positivity was required. Why Other Options Are Wrong: 5√10/2, 5/12, 12/5, 10√5: Do not satisfy both Vieta relations with the 3:2 ratio under the “m > 0” condition. Common Pitfalls: Forgetting that the sign of m depends on the sign of t via the sum relation. Also, mixing “ratio 3:2” with “difference” of roots—use the ratio parametrization directly. Final Answer: 5√10

Question 10

Roots differ by 1 (use Vieta and difference-of-roots identity): If α and β are roots of x^2 + kx + 12 = 0 with α − β = 1, find k.

Accepted Answer

Introduction / Context: For a quadratic x^2 + kx + 12 = 0, the sum and product of roots are S = −k and P = 12. The difference of roots satisfies (α − β)^2 = S^2 − 4P. Given α − β = 1, we can use this identity to determine S, and hence k, without explicitly solving for α and β. Given Data / Assumptions: S = α + β = −k P = αβ = 12 (α − β) = 1 ⇒ (α − β)^2 = 1 Concept / Approach: Use the identity (α − β)^2 = (α + β)^2 − 4αβ = S^2 − 4P. Set S^2 − 4P = 1 and solve for S. Then k = −S, which may yield two values corresponding to ±S. Step-by-Step Solution: S^2 − 4P = 1 ⇒ S^2 − 48 = 1 ⇒ S^2 = 49S = ±7 ⇒ k = −S ⇒ k = ∓7Therefore, k ∈ {7, −7} which is reported as ±7. Verification / Alternative check: Either k = 7 or k = −7 yields an equation whose roots differ by exactly 1. A quick plug with discriminant confirms that both choices produce real and distinct roots satisfying the difference condition. Why Other Options Are Wrong: ±5, ±1, 0: These do not satisfy S^2 = 49 with P = 12, hence (α − β)^2 ≠ 1. Common Pitfalls: Confusing α − β with |α − β|; here the square avoids sign issues. Also, do not attempt to solve the quadratic directly—Vieta’s identity is quicker and safer. Final Answer: ± 7

Question 11

Equal roots condition (discriminant zero): Find all k such that x^2 − 2(1 + 3k)x + 7(3 + 2k) = 0 has equal (repeated) roots.

Accepted Answer

Introduction / Context: A quadratic has equal (repeated) roots when its discriminant is zero. For ax^2 + bx + c = 0, the condition is b^2 − 4ac = 0. We apply this directly to the parameterized coefficients and solve the resulting quadratic equation in k. Given Data / Assumptions: a = 1 b = −2(1 + 3k) c = 7(3 + 2k) Equal roots ⇔ b^2 − 4ac = 0 Concept / Approach: Compute b^2 and 4ac carefully, set their difference to zero, and simplify to a standard quadratic in k. Solve that equation to obtain all admissible k values. Step-by-Step Solution: Discriminant D = [−2(1 + 3k)]^2 − 4*1*7(3 + 2k)D = 4(1 + 3k)^2 − 28(3 + 2k)Expand: 4(1 + 6k + 9k^2) − 84 − 56k = 36k^2 − 32k − 80Set D = 0 ⇒ 36k^2 − 32k − 80 = 0 ⇒ divide 4 ⇒ 9k^2 − 8k − 20 = 0k = [8 ± √(64 + 720)]/18 = [8 ± 28]/18 ⇒ k = 2 or k = −10/9 Verification / Alternative check: Substitute k = 2 and k = −10/9 back into D. In both cases, D becomes zero (by construction), confirming equal roots occur for exactly these values. Why Other Options Are Wrong: Other sign combinations do not solve 9k^2 − 8k − 20 = 0. Common Pitfalls: Arithmetic slips when expanding (1 + 3k)^2 and in combining terms with the constant 84 and 56k. Keep coefficients organized. Final Answer: 2, -10/9

Question 12

Compute λ from α^2 + β^2: If α and β are roots of x^2 − 3λx + λ^2 = 0 and α^2 + β^2 = 7/4, find λ.

Accepted Answer

Introduction / Context: Given a quadratic in parameter λ, we use Vieta’s relations to express α + β and αβ in terms of λ. Then α^2 + β^2 is computed via the identity (α + β)^2 − 2αβ. Setting this equal to the provided value yields λ directly. Given Data / Assumptions: x^2 − 3λx + λ^2 = 0 ⇒ α + β = 3λ, αβ = λ^2 α^2 + β^2 = 7/4 Concept / Approach: Use α^2 + β^2 = (α + β)^2 − 2αβ. Substitute the expressions in λ and solve the resulting equation for λ. Note that squaring creates ± solutions for λ, which are acceptable when consistent. Step-by-Step Solution: α^2 + β^2 = (3λ)^2 − 2λ^2 = 9λ^2 − 2λ^2 = 7λ^2Set 7λ^2 = 7/4 ⇒ λ^2 = 1/4 ⇒ λ = ±1/2 Verification / Alternative check: Plug λ = ±1/2 back: α + β = ±3/2, αβ = 1/4. Then α^2 + β^2 = (±3/2)^2 − 2*(1/4) = 9/4 − 1/2 = 9/4 − 2/4 = 7/4, as required. Why Other Options Are Wrong: ±√7/2, ±√3/2, ±1/3: Do not satisfy 7λ^2 = 7/4. None of these: Rejected since λ = ±1/2 works. Common Pitfalls: Forgetting the factor 2 in 2αβ or mishandling signs when squaring. Always compute α^2 + β^2 via the identity to avoid expansion errors. Final Answer: ± 1/2

Question 13

Evaluate a symmetric expression in roots: If a and b are roots of x^2 − 6x + 6 = 0, find the value of 2(a^2 + b^2).

Accepted Answer

Introduction / Context: This is a symmetric expression in the roots of a quadratic. Rather than solving for the roots explicitly, we use Vieta’s formulas together with the identity a^2 + b^2 = (a + b)^2 − 2ab to evaluate the expression in a few steps. Given Data / Assumptions: x^2 − 6x + 6 = 0 ⇒ a + b = 6, ab = 6 Compute 2(a^2 + b^2) Concept / Approach: Calculate a^2 + b^2 using the sum and product, then multiply by 2 at the end. This avoids any approximate decimal roots and keeps the computation exact and quick. Step-by-Step Solution: a^2 + b^2 = (a + b)^2 − 2ab = 6^2 − 2*6 = 36 − 12 = 242(a^2 + b^2) = 2 * 24 = 48 Verification / Alternative check: If desired, approximate roots via formula and square individually, but the identity is exact and simpler, leading to the same result 48. Why Other Options Are Wrong: 40, 42, 46, 36: These arise from arithmetic slips such as forgetting the factor 2 or miscomputing 2ab. Common Pitfalls: Mixing up ab with a^2b^2, or using (a − b)^2 instead of (a + b)^2 − 2ab. Stick to the standard identity for sum of squares. Final Answer: 48

Question 14

Express α^2/β + β^2/α in terms of coefficients: Given ax^2 + bx + c = 0 with roots α and β, find α^2/β + β^2/α in terms of a, b, c.

Accepted Answer

Introduction / Context: We are asked to write a symmetric-looking expression in the roots of a quadratic in terms of its coefficients. Vieta’s formulas give α + β = −b/a and αβ = c/a. Use algebraic identities to combine α and β into expressions involving these symmetric sums and products only. Given Data / Assumptions: ax^2 + bx + c = 0 with roots α, β S = α + β = −b/a P = αβ = c/a Target: α^2/β + β^2/α Concept / Approach: Notice α^2/β + β^2/α = (α^3 + β^3)/(αβ). Also α^3 + β^3 = (α + β)^3 − 3αβ(α + β) = S^3 − 3PS. Therefore the entire expression equals (S^3 − 3PS)/P = S^3/P − 3S. Substitute S and P in terms of a, b, c and simplify carefully. Step-by-Step Solution: S = −b/a, P = c/aS^3/P = (−b/a)^3 ÷ (c/a) = (−b^3/a^3) * (a/c) = −b^3/(a^2 c)−3S = −3(−b/a) = 3b/a = (3ab c)/(a^2 c)Sum: [−b^3 + 3abc]/(a^2 c) = (3abc − b^3)/(a^2 c) Verification / Alternative check: Test with a simple monic quadratic (a = 1) and numeric roots to confirm that both sides match numerically. The identity holds because it ultimately derives from Vieta’s symmetric relations. Why Other Options Are Wrong: They either have incorrect powers or misplace a and b in denominators, leading to dimensionally inconsistent expressions. Common Pitfalls: Dropping a power of a in the denominator or forgetting to divide by P after forming S^3 − 3PS. Keep track of a, b, c consistently. Final Answer: (3abc − b^3) / (a^2 c)

Question 15

Shift roots by +2 (build new quadratic): If α and β are roots of x^2 − 11x + 24 = 0, find the quadratic equation whose roots are (α + 2) and (β + 2).

Accepted Answer

Introduction / Context: Shifting roots by a constant k transforms the equation via the substitution y = x − k. If α, β are roots of f(y) = 0, then (α + k), (β + k) are roots of f(x − k) = 0. We use this to construct the new quadratic directly without solving for α and β numerically. Given Data / Assumptions: Original: y^2 − 11y + 24 = 0 with roots α, β We need equation with roots α + 2, β + 2 Concept / Approach: Set y = x − 2 so that when x = α + 2, we have y = α (and similarly for β). Substitute y = x − 2 into the original polynomial and expand to get the desired equation in x. Step-by-Step Solution: Start: (x − 2)^2 − 11(x − 2) + 24 = 0Expand: x^2 − 4x + 4 − 11x + 22 + 24 = 0Combine: x^2 − 15x + 50 = 0 Verification / Alternative check: Sum of new roots = (α + β) + 4 = 11 + 4 = 15 ⇒ coefficient of x is −15 (matches). Product = (α + 2)(β + 2) = αβ + 2(α + β) + 4 = 24 + 22 + 4 = 50 (constant term matches). Why Other Options Are Wrong: Each alternative has incorrect sign or constant; only x^2 − 15x + 50 = 0 matches both the shifted sum and product. Common Pitfalls: Forgetting that shifting roots adjusts both sum and product: S′ = S + 2*2 and P′ = P + 2S + 4 for a shift of +2. Final Answer: x^2 − 15x + 50 = 0

Question 16

Infinite nested radical evaluation: If x^2 = 6 + √(6 + √6 + √6 + … to infinity), find one value of x.

Accepted Answer

Introduction / Context: Infinite nested radicals often stabilize to a constant that satisfies a self-referential equation. Recognizing this allows us to solve without approximations. Here the expression under the square root repeats identically, enabling a one-variable equation for the inner radical value, and then for x. Given Data / Assumptions: x^2 = 6 + √(6 + √6 + √6 + …) Square roots are principal (nonnegative) values. Concept / Approach: Let S = √(6 + √6 + …), the infinite nested radical. Because the nesting is infinite and identical, S satisfies S = √(6 + S). Square both sides to eliminate the radical and solve for S. Then compute x from x^2 = 6 + S. Step-by-Step Solution: Let S = √(6 + S) ⇒ S^2 = 6 + SS^2 − S − 6 = 0 ⇒ (S − 3)(S + 2) = 0S ≥ 0 ⇒ S = 3 (discard S = −2)x^2 = 6 + S = 9 ⇒ x = ±3 Verification / Alternative check: Check: If x = 3, then x^2 = 9 = 6 + 3 = 6 + S, consistent. The question asks for one value; 3 is the standard principal value. Why Other Options Are Wrong: 6, 5, 4, 2: None equal a valid value of x consistent with x^2 = 9 for this setup. Common Pitfalls: Setting S = 6 + S (forgetting the square root) or using S = 6 + √S. The correct self-reference is S = √(6 + S). Final Answer: 3

Question 17

Algebraic simplification (compound fraction): If a = p/(p + q) and b = q/(p − q), evaluate ab / (a + b) in simplest form.

Accepted Answer

Introduction / Context: This problem requires careful manipulation of rational expressions. By putting a and b over a common denominator, we can form a + b and ab explicitly and then simplify the compound fraction ab/(a + b). The key move is recognizing that identical factors cancel cleanly. Given Data / Assumptions: a = p/(p + q) b = q/(p − q) p ≠ ±q (to avoid zero denominators) Concept / Approach: Compute numerator and denominator separately: ab and a + b. Use the identity (p + q)(p − q) = p^2 − q^2 and simplify. The cancellation of (p^2 − q^2) between numerator and denominator will leave a compact expression in p and q. Step-by-Step Solution: ab = [p/(p + q)] * [q/(p − q)] = pq / (p^2 − q^2)a + b = p/(p + q) + q/(p − q) = [p(p − q) + q(p + q)] / (p^2 − q^2)Simplify numerator: p^2 − pq + qp + q^2 = p^2 + q^2Thus ab/(a + b) = [pq/(p^2 − q^2)] / [(p^2 + q^2)/(p^2 − q^2)] = pq/(p^2 + q^2) Verification / Alternative check: Test with concrete values (e.g., p = 3, q = 1) to ensure the simplified expression matches the numeric computation of ab/(a + b). It does. Why Other Options Are Wrong: Other options either invert the result or drop necessary terms, failing to account for the combination in the denominator. Common Pitfalls: Forgetting to combine like terms p(−q) + q(p) = 0 (they cancel), leaving p^2 + q^2. Also, do not try to add numerators without establishing the common denominator first. Final Answer: pq / (p^2 + q^2)

Question 18

Sum of roots equals sum of squares: For x^2 + px + q = 0, if (sum of roots) = (sum of squares of roots), find the relation between p and q.

Accepted Answer

Introduction / Context: Using Vieta’s formulas, we can relate the sum and product of roots of a quadratic to its coefficients. The condition equating the sum of the roots to the sum of their squares produces an algebraic relation between p and q, which we simplify to the requested form. Given Data / Assumptions: Equation: x^2 + px + q = 0 Sum of roots S = −p Product P = q Condition: S = (sum of squares) = S^2 − 2P Concept / Approach: Compute the sum of squares via S^2 − 2P, set equal to S, and rearrange to isolate p, q in a neat relation. This is a classic symmetric manipulation avoiding any need to find individual roots. Step-by-Step Solution: S = S^2 − 2P ⇒ −p = p^2 − 2qRearrange: p^2 + p − 2q = 0 ⇒ p^2 + p = 2q Verification / Alternative check: Choose a numeric pair (p, q) satisfying p^2 + p = 2q, generate the quadratic, compute its roots, and verify that r1 + r2 equals r1^2 + r2^2. The identity is exact. Why Other Options Are Wrong: Other relations do not follow from S = S^2 − 2P and conflict with Vieta’s formulas. Common Pitfalls: Sign errors when substituting S = −p; remember P = q. Do not confuse S^2 − 2P with (r1 − r2)^2. Final Answer: p^2 + p = 2q

Question 19

Evaluate a cube-root symmetric sum: If α and β are roots of 8x^2 − 3x + 27 = 0, find (α^2/β)^{1/3} + (β^2/α)^{1/3}.

Accepted Answer

Introduction / Context: This is a nonstandard but elegant manipulation with roots of a quadratic. Define A = (α^2/β)^{1/3} and B = (β^2/α)^{1/3}. We are to find S = A + B. Using identities for A^3 + B^3 and AB, we can form a cubic in S and then locate the rational value among the options that satisfies it. Given Data / Assumptions: α + β = 3/8 (from −b/a) αβ = 27/8 (from c/a) A^3 = α^2/β, B^3 = β^2/α AB = (αβ)^{1/3} Concept / Approach: We use (A + B)^3 = A^3 + B^3 + 3AB(A + B). Compute A^3 + B^3 in terms of α, β using the identity α^3 + β^3 = (α + β)^3 − 3αβ(α + β), then divide by αβ. Compute AB directly as (αβ)^{1/3}. Substitute into the cubic for S and solve by testing the given rational choices. Step-by-Step Solution: Let S1 = α + β = 3/8, P = αβ = 27/8α^3 + β^3 = S1^3 − 3PS1So A^3 + B^3 = (α^3 + β^3)/P = (S1^3 − 3PS1)/PCompute: S1^3 = 27/512; 3PS1 = 3*(27/8)*(3/8) = 243/64Hence A^3 + B^3 = (27/512 − 1944/512)/(27/8) = (−1917/512)*(8/27) = −71/64AB = P^{1/3} = (27/8)^{1/3} = 3/2Let S = A + B. Then S^3 = (−71/64) + 3*(3/2)*S = (−71/64) + (9/2)SMultiply by 64: 64S^3 − 288S + 71 = 0Test options: S = 1/4 satisfies 64*(1/64) − 288*(1/4) + 71 = 1 − 72 + 71 = 0 Verification / Alternative check: None of the other options satisfy the cubic. Since the cubic has S = 1/4 as a root and other options are far off, 1/4 is the correct value. Why Other Options Are Wrong: 1/3, 7/2, 4, 3/4: Substitution into 64S^3 − 288S + 71 fails to give zero. Common Pitfalls: Arithmetic slips when converting to a common denominator or when dividing by P. Keep fractions exact; avoid decimals to prevent rounding errors. Final Answer: 1/4

Question 20

Back out the correct quadratic from two different mistakes: One student mis-copies the coefficient of x and gets roots −9 and −1. Another mis-copies the constant term and gets roots 8 and 2. Find the correct quadratic equation.

Accepted Answer

Introduction / Context: This problem describes two different erroneous versions of the same intended quadratic. One student changed only the x-coefficient, the other only the constant term. Using the sums and products of the resulting roots, we can reconstruct the correct coefficients that are consistent with both stories. Given Data / Assumptions: Correct equation is monic: x^2 + Bx + C = 0 Wrong-x-coefficient student obtains roots −9 and −1 ⇒ sum −10, product 9 Wrong-constant student obtains roots 8 and 2 ⇒ sum 10, product 16 Each student changes only one coefficient (B or C) relative to the correct equation Concept / Approach: From the first student: product equals the correct C, because only B changed. So C = 9. From the second student: sum equals the negative of the correct B, because only C changed. Since 8 + 2 = 10, we get −B = 10 ⇒ B = −10. Thus the correct equation is x^2 − 10x + 9 = 0. Step-by-Step Solution: First student: product = C = (−9)(−1) = 9Second student: sum = −B = 8 + 2 = 10 ⇒ B = −10Correct equation: x^2 − 10x + 9 = 0 Verification / Alternative check: Check each wrong case: With C fixed at 9, choose B′ = +10 to get roots −9 and −1. With B fixed at −10, choose C′ = 16 to get roots 8 and 2. Both match the described mistakes. Why Other Options Are Wrong: x^2 + 10x + 9 and x^2 − 10x + 16 conflict with the deduced B and C. None of the above: Not applicable since x^2 − 10x + 9 works. Common Pitfalls: Mixing up which coefficient stays the same in each erroneous equation. Remember: changing the x-coefficient alters the sum; changing the constant alters the product. Final Answer: x^2 − 10x + 9 = 0

Quadratic Equation Questions