Question 1

A rectangular plot has length 8 m greater than its breadth. If its area is 308 m², find the length of the plot.

Accepted Answer

Introduction / Context: Relate dimensions of a rectangle using algebra: let breadth be b (in meters). Given that length is b + 8 and the area is 308, form a quadratic in b and solve. Then compute the length from b + 8. Given Data / Assumptions: Length L = b + 8 Area A = 308 m^2 Area formula: A = L * b Concept / Approach: Set up b(b + 8) = 308 to obtain a quadratic. Solve for b, discard the negative root as a physical dimension, and add 8 to get L. Step-by-Step Solution: b(b + 8) = 308 ⇒ b^2 + 8b − 308 = 0Discriminant: 8^2 + 4*308 = 64 + 1232 = 1296 ⇒ √1296 = 36b = (−8 ± 36)/2 ⇒ b = 14 or b = −22 (reject)Length L = b + 8 = 14 + 8 = 22 m Verification / Alternative check: Area check: 22 * 14 = 308 m^2, matches the given area. Why Other Options Are Wrong: 18 m and 20 m do not produce area 308 with an 8 m difference; “None of these” is unnecessary since 22 m fits perfectly; 24 m would imply breadth 16 m and area 384 m^2. Common Pitfalls: Arithmetic mistakes with the discriminant or selecting the negative breadth. Ensure units in m^2 for area. Final Answer: 22 m

Question 2

Solve for x: (x + 1)^2 − 3(x − 1) = 4. Choose the correct value of x.

Accepted Answer

Introduction / Context: This is a straightforward quadratic simplification leading to a linear factorization. Expand the square, distribute the subtraction, and collect like terms to solve for x. The options include simple rational or integer candidates. Given Data / Assumptions: Equation: (x + 1)^2 − 3(x − 1) = 4 All real solutions are admissible. Concept / Approach: Expand (x + 1)^2, distribute −3 across (x − 1), then move all terms to one side to get a factorable expression. Solve the resulting equation and pick the matching option. Step-by-Step Solution: Expand: (x + 1)^2 = x^2 + 2x + 1Distribute: −3(x − 1) = −3x + 3Sum: x^2 + 2x + 1 − 3x + 3 = 4 ⇒ x^2 − x + 4 = 4Simplify: x^2 − x = 0 ⇒ x(x − 1) = 0Solutions: x = 0 or x = 1 Verification / Alternative check: Plug x = 0: LHS = 1 − (−3) = 4 = RHS. Plug x = 1: LHS = 4 − 0 = 4. Both work; among the listed options, 0 is explicitly present. Why Other Options Are Wrong: −2, 1/2, −1 are not solutions of x(x − 1) = 0; 1 is also a valid solution but not marked correct here since the single correct choice is specified as 0. Common Pitfalls: Algebraic slips when expanding or moving the 4; forgetting there can be two solutions before matching with discrete options. Final Answer: 0

Question 3

If x^2 − 3x + 1 = 0, compute the value of x + 1/x.

Accepted Answer

Introduction / Context: When x satisfies a quadratic, expressions like x + 1/x can be simplified by using the quadratic to eliminate x^2. This avoids explicitly solving for x and keeps the algebra light and clean. Given Data / Assumptions: x^2 − 3x + 1 = 0 x ≠ 0 (since 1/x is used) Concept / Approach: Compute x + 1/x = (x^2 + 1)/x. Replace x^2 using the given quadratic: x^2 = 3x − 1. This method is standard for symmetric expressions in x and 1/x. Step-by-Step Solution: x + 1/x = (x^2 + 1)/xGiven x^2 = 3x − 1Hence (x^2 + 1)/x = (3x − 1 + 1)/x = 3x/x = 3 Verification / Alternative check: Alternatively, divide the original equation by x: x − 3 + 1/x = 0 ⇒ x + 1/x = 3, same result. Why Other Options Are Wrong: 0, 2, 1, −1 do not match the identity derived directly from the quadratic. Common Pitfalls: Forgetting to ensure x ≠ 0 (it is not a root here), or mishandling algebra when substituting x^2. Final Answer: 3

Question 4

If one root of 7y^2 − 50y + k = 0 is y = 7, find the value of k.

Accepted Answer

Introduction / Context: A known root of a polynomial must zero the polynomial when substituted. Use this to solve for the unknown parameter k quickly by plugging y = 7 into the given quadratic expression. Given Data / Assumptions: Quadratic: 7y^2 − 50y + k = 0 Root: y = 7 Concept / Approach: Direct substitution: set 7*(7^2) − 50*7 + k = 0 and solve for k. Keep arithmetic neat to avoid sign errors. Step-by-Step Solution: Compute: 7*(49) − 50*7 + k = 0343 − 350 + k = 0 ⇒ −7 + k = 0Therefore, k = 7 Verification / Alternative check: Substitute back: 7*(49) − 350 + 7 = 343 − 343 = 0, confirming y = 7 is indeed a root. Why Other Options Are Wrong: 1, 50/7, 7/50, 0 do not satisfy the zero condition upon substitution. Common Pitfalls: Arithmetic slips when multiplying 50*7 or 7*49. Double-check the sign before solving. Final Answer: 7

Question 5

Find k so that x^2 + 2(k − 4)x + 2k = 0 has equal (repeated) roots. Determine the value(s) of k.

Accepted Answer

Introduction / Context: A quadratic has equal (repeated) real roots when its discriminant is zero. Apply this criterion to the parameterized quadratic x^2 + 2(k − 4)x + 2k = 0 to find all k that force equality of roots. Given Data / Assumptions: Quadratic: x^2 + 2(k − 4)x + 2k = 0 Equal roots ⇔ Discriminant D = 0 Concept / Approach: Compute D = [2(k − 4)]^2 − 4*1*(2k). Set D = 0 and solve the resulting quadratic in k. Both solutions (if real) are acceptable parameter values. Step-by-Step Solution: D = 4(k − 4)^2 − 8k = 0Divide by 4: (k − 4)^2 − 2k = 0Expand: k^2 − 8k + 16 − 2k = 0 ⇒ k^2 − 10k + 16 = 0Solve: k = [10 ± √(100 − 64)]/2 = (10 ± 6)/2 ⇒ k = 8 or k = 2 Verification / Alternative check: For k = 8: equation is x^2 + 8x + 16 = 0 ⇒ (x + 4)^2. For k = 2: x^2 − 4x + 4 = 0 ⇒ (x − 2)^2. Both have equal roots. Why Other Options Are Wrong: Pairs like “6 and 4”, “10 and 4”, “12 and 2” do not satisfy D = 0 for both k values; “Only 8” ignores k = 2, which also works. Common Pitfalls: Forgetting to divide by 4 when simplifying D or making sign errors when expanding (k − 4)^2. Always verify by completing the square. Final Answer: 8 and 2

Question 6

Quadratic construction: Find the monic quadratic equation whose real roots are 3 and −1. Choose the correct standard-form equation.

Accepted Answer

Introduction / Context: The question asks you to build a quadratic equation from its roots. This is a standard skill in algebra: once the roots are known, the quadratic can be written quickly using the factor form and then expanded to standard form ax^2 + bx + c = 0. Given Data / Assumptions: Roots: r1 = 3 and r2 = −1. We are to write a monic quadratic (leading coefficient 1). Standard form required for comparison with options. Concept / Approach: If r1 and r2 are roots, the monic polynomial is (x − r1)(x − r2) = 0. Expand the product to get coefficients. Alternatively, use sum and product of roots: sum = r1 + r2, product = r1 * r2. Then x^2 − (sum)x + (product) = 0. Step-by-Step Solution: Sum of roots S = 3 + (−1) = 2.Product of roots P = 3 * (−1) = −3.Monic quadratic: x^2 − Sx + P = x^2 − 2x − 3 = 0. Verification / Alternative check: Factor x^2 − 2x − 3 = (x − 3)(x + 1) = 0 which gives roots 3 and −1 as required. Why Other Options Are Wrong: Each incorrect option has either the wrong sign for x-term or constant term, yielding different roots upon factoring. Common Pitfalls: Mixing up signs when forming x^2 − (sum)x + (product); always compute sum and product carefully. Final Answer: x^2 − 2x − 3 = 0

Question 7

Root transformation: If α and β are the roots of 4x^2 − 19x + 12 = 0, find the quadratic equation whose roots are 1/α and 1/β.

Accepted Answer

Introduction / Context: This tests the reciprocal-root transformation. Given a quadratic with roots α and β, we are to construct the equation for roots 1/α and 1/β without explicitly finding α and β. Given Data / Assumptions: Original equation: 4x^2 − 19x + 12 = 0 with roots α, β. We want the equation with roots 1/α and 1/β. Coefficients are real and the quadratic is non-degenerate. Concept / Approach: For ax^2 + bx + c = 0, sum α + β = −b/a and product αβ = c/a. Then for reciprocals: (1/α) + (1/β) = (α + β)/(αβ) and (1/α)(1/β) = 1/(αβ). The monic form is x^2 − [(α + β)/(αβ)]x + [1/(αβ)] = 0, which can be scaled to clear denominators. Step-by-Step Solution: a = 4, b = −19, c = 12.α + β = −b/a = 19/4; αβ = c/a = 12/4 = 3.For reciprocals: sum = (α + β)/(αβ) = (19/4)/3 = 19/12; product = 1/(αβ) = 1/3.Equation: x^2 − (19/12)x + 1/3 = 0 → multiply by 12: 12x^2 − 19x + 4 = 0. Verification / Alternative check: Direct transformation rule: cx^2 + bx + a = 0 for reciprocals gives 12x^2 − 19x + 4 = 0, consistent. Why Other Options Are Wrong: Signs or placements of coefficients are incorrect; only option with pattern c, b, a works. Common Pitfalls: Forgetting to divide by product αβ when summing reciprocals; not clearing denominators cleanly. Final Answer: 12x^2 − 19x + 4 = 0

Question 8

Compute the absolute difference between the two real roots of the quadratic equation 2x^2 − 11x + 5 = 0.

Accepted Answer

Introduction / Context: The distance between the roots of a quadratic can be computed without solving for each root explicitly by using the identity |α − β| = sqrt((α + β)^2 − 4αβ). This leverages sum and product of roots from coefficients. Given Data / Assumptions: Equation: 2x^2 − 11x + 5 = 0. Real roots exist (discriminant positive). Concept / Approach: For ax^2 + bx + c = 0, α + β = −b/a and αβ = c/a. Then |α − β| = sqrt((α + β)^2 − 4αβ). This avoids computing α and β separately. Step-by-Step Solution: a = 2, b = −11, c = 5.α + β = −b/a = 11/2.αβ = c/a = 5/2.|α − β| = sqrt((11/2)^2 − 4*(5/2)) = sqrt(121/4 − 10) = sqrt(81/4) = 9/2 = 4.5. Verification / Alternative check: Solving explicitly via quadratic formula also yields roots whose difference is 4.5; the identity is quicker and exact. Why Other Options Are Wrong: They correspond to common arithmetic slips (e.g., missing the division by a or miscomputing the constant term). Common Pitfalls: Using sqrt(b^2 − 4ac)/a directly and forgetting to divide by |a|, which leads to the same value but often mishandled. Final Answer: 4.5

Question 9

Build the quadratic equation whose solution set is {2, 1/4}. Give the equation in integer-coefficient standard form.

Accepted Answer

Introduction / Context: Constructing a quadratic from given roots is foundational. When roots are fractional, it is good practice to clear denominators to present an equation with integer coefficients. Given Data / Assumptions: Roots: 2 and 1/4. We want a standard form with integer coefficients. Concept / Approach: Use x^2 − (sum)x + (product) = 0. Compute sum and product carefully with the fraction, then multiply by the least common multiple of denominators to clear fractions. Step-by-Step Solution: Sum S = 2 + 1/4 = 9/4.Product P = 2 * 1/4 = 1/2.Monic: x^2 − (9/4)x + 1/2 = 0.Clear denominators by multiplying by 4: 4x^2 − 9x + 2 = 0. Verification / Alternative check: Factor check: (4x − 1)(x − 2) = 4x^2 − 8x − x + 2 = 4x^2 − 9x + 2, so roots are 1/4 and 2. Why Other Options Are Wrong: They represent incorrect arithmetic (wrong sum/product) or not both required roots. Common Pitfalls: Forgetting to clear denominators or mixing signs in the sum term. Final Answer: 4x^2 − 9x + 2 = 0

Question 10

Let p and q be the roots of x^2 + p x + q = 0. Determine the correct values of (p, q) consistent with this definition.

Accepted Answer

Introduction / Context: Here the symbolic coefficients p and q also equal the roots of their own quadratic. This creates equations among p and q via Vieta’s relations that must be satisfied simultaneously. Given Data / Assumptions: Equation: x^2 + px + q = 0. Roots are p and q themselves. Real values expected. Concept / Approach: For monic x^2 + px + q = 0 with roots r1 = p and r2 = q: sum r1 + r2 = −p and product r1*r2 = q (Vieta). This yields a system in p and q. Step-by-Step Solution: Sum: p + q = −p → q = −2p.Product: pq = q.Either q = 0 or p = 1.If q = 0 then q = −2p → p = 0 (gives (0, 0), not in options). If p = 1 then q = −2.Therefore (p, q) = (1, −2). Verification / Alternative check: With p = 1, q = −2, the quadratic x^2 + x − 2 = 0 has roots 1 and −2, matching p and q. Why Other Options Are Wrong: They violate either the sum or product condition; check Vieta’s relations to see inconsistencies. Common Pitfalls: Missing the special solution q = 0, p = 0 (not listed) or confusing coefficient p with root p without applying Vieta correctly. Final Answer: p = 1 and q = −2

Question 11

Solve for the number of real solutions of the equation: √(x^2 − x + 1) + 1/√(x^2 − x + 1) = 2 − x^2.

Accepted Answer

Introduction / Context: This equation mixes a radical expression and its reciprocal on the left with a quadratic on the right. An inequality insight about t + 1/t helps bound the left-hand side and identify feasible x-values without brute force algebra. Given Data / Assumptions: A(x) = x^2 − x + 1. LHS: √A + 1/√A, domain requires A > 0. RHS: 2 − x^2. Concept / Approach: Let t = √A > 0. Then LHS = t + 1/t ≥ 2 by AM ≥ GM, with equality iff t = 1. Also, A = x^2 − x + 1 has minimum 3/4 (at x = 1/2), so t ≥ √(3/4) = √3/2 > 0, ensuring the expression is defined for all real x. Step-by-Step Solution: Since t + 1/t ≥ 2, LHS ≥ 2, with equality when t = 1 → A = 1.A = 1 ⇒ x^2 − x + 1 = 1 ⇒ x(x − 1) = 0 ⇒ x = 0 or x = 1.Check RHS at these x: for x = 0, RHS = 2 − 0 = 2; LHS = 1 + 1 = 2 → valid.For x = 1, RHS = 2 − 1 = 1; LHS = 2 → not valid.For any other x, LHS > 2 while RHS ≤ 2, so no equality. Verification / Alternative check: Numerical sampling near x = 0 and x = 1 confirms only x = 0 satisfies the equality. Why Other Options Are Wrong: 0, 2, 3, 4 solutions contradict the inequality structure; only one real solution exists. Common Pitfalls: Forgetting the equality condition t = 1 for t + 1/t = 2; mishandling the domain by allowing A ≤ 0 (which never occurs here). Final Answer: 1

Question 12

Two numbers have sum 8 and difference 4. Form the quadratic equation having these numbers as roots, and pick the correct option.

Accepted Answer

Introduction / Context: From the sum and difference of two numbers, you can determine the numbers and then the quadratic having them as roots. This illustrates reverse construction of a polynomial from root properties. Given Data / Assumptions: Sum S = 8. Difference D = 4 (so numbers are (S ± D)/2). Quadratic desired with those numbers as roots. Concept / Approach: Compute the two numbers, then use x^2 − Sx + P = 0, where P is their product. Alternatively, directly use S and P without computing the numbers individually. Step-by-Step Solution: Numbers: (8 ± 4)/2 ⇒ 6 and 2.Sum S = 8, Product P = 6 * 2 = 12.Quadratic: x^2 − Sx + P = x^2 − 8x + 12 = 0. Verification / Alternative check: Factor: (x − 6)(x − 2) = x^2 − 8x + 12 ⇒ roots 6 and 2, consistent. Why Other Options Are Wrong: Sign errors in the x-term or constant term change the sum/product, creating different roots. Common Pitfalls: Taking D itself as a root or misusing S and P in the formula. Final Answer: x^2 − 8x + 12 = 0

Question 13

If a number and its reciprocal have a sum of 10/3, determine the number(s).

Accepted Answer

Introduction / Context: Sum-with-reciprocal problems lead to a quadratic after clearing denominators. Solving that quadratic yields the pair of numbers, which are mutual reciprocals by construction. Given Data / Assumptions: x + 1/x = 10/3. x ≠ 0. Concept / Approach: Multiply by x to obtain a standard quadratic in x. Solve using the quadratic formula or factoring if the discriminant is a perfect square. Step-by-Step Solution: x + 1/x = 10/3 ⇒ multiply both sides by x: x^2 + 1 = (10/3)x.Rearrange: 3x^2 − 10x + 3 = 0.Discriminant Δ = (−10)^2 − 4*3*3 = 100 − 36 = 64.Roots: x = (10 ± 8)/6 ⇒ 18/6 = 3 or 2/6 = 1/3. Verification / Alternative check: Check: 3 + 1/3 = 10/3; also 1/3 + 3 = 10/3, both valid. Why Other Options Are Wrong: Options with negative signs do not satisfy the given positive sum; “10/3 only” is not a number satisfying x + 1/x = 10/3. Common Pitfalls: Forgetting to multiply through by x or mishandling the quadratic coefficients. Final Answer: 3 and 1/3

Question 14

Find all real values of p for which the difference between the roots of x^2 + p x + 8 = 0 equals 2.

Accepted Answer

Introduction / Context: The spacing of roots of a monic quadratic can be related directly to its coefficients using Vieta’s relations and the identity |α − β| = sqrt((α + β)^2 − 4αβ). Here, the spacing is fixed at 2; we solve for the parameter p that enforces this. Given Data / Assumptions: Equation: x^2 + p x + 8 = 0. |α − β| = 2. We seek real p values. Concept / Approach: For monic quadratics, α + β = −p and αβ = 8. Then |α − β| = sqrt((α + β)^2 − 4αβ) = sqrt(p^2 − 32). Set this equal to 2 and solve. Step-by-Step Solution: sqrt(p^2 − 32) = 2 ⇒ p^2 − 32 = 2^2 = 4.p^2 = 36 ⇒ p = ±6. Verification / Alternative check: With p = 6 or p = −6, compute the discriminant and verify the roots’ difference equals 2; both values work. Why Other Options Are Wrong: ±2, ±4, ±8 do not satisfy p^2 − 32 = 4; “No real p” contradicts the solution. Common Pitfalls: Forgetting the absolute value in the difference or squaring both sides incorrectly. Final Answer: p = ±6

Question 15

Let x be the greater real root of x^2 + x − 20 = 0 and y be the greater real root of y^2 − y − 30 = 0. Compare x and y and choose the correct relation.

Accepted Answer

Introduction / Context: The original statement was ambiguous because each quadratic has two roots. Applying the Recovery-First Policy, we define x and y explicitly as the greater real roots. We then compare these values by solving each quadratic via factoring. Given Data / Assumptions: x solves x^2 + x − 20 = 0 and is the greater root. y solves y^2 − y − 30 = 0 and is the greater root. We compare x to y. Concept / Approach: Factor each quadratic to get roots quickly. For ax^2 + bx + c with small integers, try integer factor pairs of c that sum to b. Identify the larger root in each case and compare. Step-by-Step Solution: x^2 + x − 20 = 0 ⇒ (x + 5)(x − 4) = 0 ⇒ roots: −5, 4 ⇒ greater x = 4.y^2 − y − 30 = 0 ⇒ (y − 6)(y + 5) = 0 ⇒ roots: 6, −5 ⇒ greater y = 6.Therefore x = 4 and y = 6 ⇒ x < y. Verification / Alternative check: Quadratic formula gives identical results; factoring is sufficient here. Why Other Options Are Wrong: x > y or x ≥ y contradict the computed values; equality does not hold. Common Pitfalls: Not clarifying which root is used; comparing mismatched (e.g., larger with smaller) roots gives inconsistent relations. Final Answer: If x < y

Question 16

Let x be the greater real root of 225x^2 − 4 = 0 and y be the (unique) real solution of 15y + 2 = 0 (interpreting √225 · y + 2 = 0). Compare x and y.

Accepted Answer

Introduction / Context: The second relation was typographically unclear. Using Recovery-First Policy, we interpret √225 y + 2 = 0 as (√225)·y + 2 = 0 ⇒ 15y + 2 = 0. We then solve both and compare the numeric values using the greater real root convention for the quadratic. Given Data / Assumptions: 225x^2 − 4 = 0 ⇒ x has two symmetric real roots; take the greater one. 15y + 2 = 0 ⇒ single solution y. Concept / Approach: Solve each: isolate x^2 and y. For comparison, evaluate numerically and select the correct inequality. Step-by-Step Solution: 225x^2 − 4 = 0 ⇒ x^2 = 4/225 ⇒ x = ±2/15; greater root x = 2/15.15y + 2 = 0 ⇒ y = −2/15.Compare: 2/15 > −2/15 ⇒ x > y. Verification / Alternative check: Decimal check: x ≈ 0.133…, y ≈ −0.133…, confirms the inequality. Why Other Options Are Wrong: They reverse the sign relation or claim equality, which is false given the computed values. Common Pitfalls: Misreading √225 y as √(225y); the intended linear form is (√225)·y. Final Answer: If x > y

Question 17

Let x satisfy x − √121 = 0 and y be the greater real root of y^2 − 121 = 0. Compare x and y.

Accepted Answer

Introduction / Context: We compare specific solutions: x is uniquely determined by a linear radical equation; y comes from a quadratic with two symmetric roots. By convention, we take the greater real root for y to avoid ambiguity and then compare numerically. Given Data / Assumptions: x − √121 = 0 ⇒ x = √121 = 11. y^2 − 121 = 0 ⇒ y = ±11, take greater y = 11. Concept / Approach: Evaluate both exactly, then choose the correct relation symbol. Step-by-Step Solution: x = 11.y (greater root) = 11.Thus x = y, which satisfies x ≥ y (and also x ≤ y). The offered relation set includes ≥. Verification / Alternative check: Direct substitution confirms equality. Why Other Options Are Wrong: Strict inequalities (x > y or x < y) are false because the values are equal. Common Pitfalls: Forgetting to select the greater root for y or mis-evaluating √121. Final Answer: If x ≥ y

Question 18

Let x be the greater real root of x^2 − 16 = 0 and y be the greater real root of y^2 − 9y + 20 = 0. Compare x and y.

Accepted Answer

Introduction / Context: Two quadratics with easily factorable forms allow immediate determination of greater roots. We then compare those values directly. Given Data / Assumptions: x^2 − 16 = 0 ⇒ roots ±4, greater x = 4. y^2 − 9y + 20 = 0 ⇒ (y − 4)(y − 5) = 0 ⇒ roots 4 and 5, greater y = 5. Concept / Approach: Factorization and identification of the larger root for each equation. Step-by-Step Solution: x = 4 (greater root of first).y = 5 (greater root of second).Therefore x < y. Verification / Alternative check: Numeric comparison is straightforward; no alternative method needed. Why Other Options Are Wrong: They contradict the ordering 4 < 5. Common Pitfalls: Accidentally picking the smaller root 4 for y instead of 5; the prompt specifies the greater root. Final Answer: if x < y

Question 19

Let x be the greater real root of x^2 − 32 = 112 and y satisfy y − √169 = 0. Compare x and y.

Accepted Answer

Introduction / Context: One equation is a shifted square giving an exact integer root; the other is a simple radical linear relation. We compute both values and compare using the greater-root convention where applicable. Given Data / Assumptions: x^2 − 32 = 112 ⇒ x^2 = 144 ⇒ roots ±12; greater x = 12. y − √169 = 0 ⇒ y = √169 = 13. Concept / Approach: Direct evaluation suffices. Step-by-Step Solution: x = 12 (greater root).y = 13.Hence x < y. Verification / Alternative check: 12 vs 13 confirms the inequality. Why Other Options Are Wrong: They contradict the established ordering. Common Pitfalls: Mis-evaluating √169 as something other than 13 or forgetting the “greater root” constraint for x. Final Answer: If x < y

Question 20

Let x be the greater real root of x^2 − 8x + 15 = 0 and y be the greater real root of y^2 − 3y + 2 = 0. Compare x and y.

Accepted Answer

Introduction / Context: Both quadratics factor neatly. After identifying the larger root in each case, a simple comparison yields the correct relation between x and y. Given Data / Assumptions: x^2 − 8x + 15 = 0 ⇒ (x − 3)(x − 5) = 0 ⇒ greater x = 5. y^2 − 3y + 2 = 0 ⇒ (y − 1)(y − 2) = 0 ⇒ greater y = 2. Concept / Approach: Factor, select the greater root, compare numerically. Step-by-Step Solution: x = 5; y = 2.Therefore x > y. Verification / Alternative check: The quadratic formula yields identical values; factoring is quickest. Why Other Options Are Wrong: They contradict the numerical ordering 5 > 2. Common Pitfalls: Accidentally picking the smaller root in either quadratic. Final Answer: If x > y

Quadratic Equation Questions