Question 1

Identify the quadratic equation: Which of the following is a quadratic equation (an equation of degree 2)?

Accepted Answer

Introduction / Context:A quadratic equation is any equation that can be written in the standard form ax^2 + bx + c = 0 with a ≠ 0. We examine each option for its degree after arranging it as an equation (if needed) to decide which one is quadratic. Given Data / Assumptions: Degree means the highest power of x in the equation after simplification. We can rearrange expressions like “x^4 − 10” to an equation by setting them equal to zero if intended. Concept / Approach:Check each option’s degree. Quadratic ⇔ degree 2 once terms are brought to one side. Anything of degree 3, 4, or higher (or not an equation) is not quadratic. Step-by-Step Solution: Option (c): 7x^2 = 49 ⇒ 7x^2 − 49 = 0 ⇒ degree 2 (quadratic)(a): degree 3; (b): degree 4; (d): degree 4; (e): degree 5 Verification / Alternative check:Solving (c) gives x^2 = 7 ⇒ real solutions x = ±√7, consistent with a quadratic form. Why Other Options Are Wrong: (a), (d), (e): Not degree 2; they are cubic, quartic, and quintic respectively. (b): Also degree 4 when written as x^4 − 10 = 0. Common Pitfalls:Confusing “contains x^2” with “is quadratic.” Presence of higher powers (x^3, x^4, etc.) means the equation is not quadratic. Final Answer:7x^2 = 49

Question 2

Which equation has real roots? Among the following equations, identify the one that definitely has real roots.

Accepted Answer

Introduction / Context:To decide whether a quadratic has real roots, we can use factoring or the discriminant test (b^2 − 4ac). Factored forms immediately reveal roots; otherwise, compute the discriminant to infer the nature of roots. Given Data / Assumptions: We examine each option for factorization or discriminant sign. Real roots exist if discriminant ≥ 0. Concept / Approach:Option (b) is already factored into two linear terms, giving real roots directly. For (a) and (c), we check discriminants. Option (d) is not an equation and is irrelevant as an answer. Option (e) is a known quadratic with negative discriminant. Step-by-Step Solution: (b): (x − 1)(2x − 5) = 0 ⇒ roots x = 1 and x = 2.5 (real)(a): 2x^2 − 3x + 4 ⇒ D = (−3)^2 − 4*2*4 = 9 − 32 = −23 (no real roots)(c): 3x^2 + 4x + 5 ⇒ D = 4^2 − 4*3*5 = 16 − 60 = −44 (no real roots)(e): x^2 + x + 1 ⇒ D = 1 − 4 = −3 (no real roots) Verification / Alternative check:Factored forms guarantee real roots because each linear factor equals zero at a real x-value. Why Other Options Are Wrong: (a), (c), (e): Negative discriminants imply complex (non-real) roots. (d): Not a mathematical equation; cannot be “real roots.” Common Pitfalls:Ignoring signs in b^2 − 4ac or miscomputing products. Keep coefficients clear. Final Answer:(x − 1)(2x − 5) = 0

Question 3

Solve the quadratic completely: Find the roots of 2x^2 − 9x − 18 = 0.

Accepted Answer

Introduction / Context:We solve a quadratic equation using the quadratic formula or by factoring if possible. The discriminant indicates whether the roots are real and helps simplify radicals when they are perfect squares. Given Data / Assumptions: Equation: 2x^2 − 9x − 18 = 0 a = 2, b = −9, c = −18 Concept / Approach:Use the quadratic formula x = [−b ± √(b^2 − 4ac)]/(2a). Since the discriminant is a perfect square, the roots will be rational and can be presented in fractional form. Step-by-Step Solution: D = b^2 − 4ac = (−9)^2 − 4*2*(−18) = 81 + 144 = 225√D = √225 = 15x = [9 ± 15]/(2*2) = (9 ± 15)/4Roots: x = 24/4 = 6 and x = −6/4 = −3/2 Verification / Alternative check:Substitute x = 6: 2*36 − 54 − 18 = 0. Substitute x = −3/2: 2*(9/4) + (27/2) − 18 = 0. Both satisfy the equation. Why Other Options Are Wrong: 3/2 and 6; 3/2 and −6; −3/2 and −6: Each pair has at least one sign incorrect. −6 and 6: −6 is not a root; −3/2 is. Common Pitfalls:Dropping the negative sign on b or mishandling the division by 2a when simplifying. Final Answer:-3/2 and 6

Question 4

Parameter from a known root: If one root of x^2 − 6kx + 5 = 0 is 5, find the value of k.

Accepted Answer

Introduction / Context:When a root of a polynomial is known, substituting it into the equation yields a direct relation among parameters. Here, plugging x = 5 into the quadratic allows us to solve for k in one step. Given Data / Assumptions: Equation: x^2 − 6kx + 5 = 0 One root is x = 5 Concept / Approach:If r is a root, then r satisfies the equation identically. Substitute r = 5 and solve the resulting linear equation for k. No need for the quadratic formula here. Step-by-Step Solution: 5^2 − 6k*5 + 5 = 025 − 30k + 5 = 0 ⇒ 30 − 30k = 0k = 1 Verification / Alternative check:With k = 1, equation becomes x^2 − 6x + 5 = 0, which factors as (x − 1)(x − 5) = 0, confirming 5 is indeed a root. Why Other Options Are Wrong: −1/2, −1, 2, 0: Substitution does not result in zero; they fail the root condition for x = 5. Common Pitfalls:Forgetting to multiply 6k by x or mishandling algebraic signs when combining constants. Final Answer:1

Question 5

Find the roots exactly: Solve 2x^2 − 11x + 15 = 0.

Accepted Answer

Introduction / Context:Factoring is often the fastest way to solve a quadratic with integer coefficients if a suitable factor pair exists. Here the numbers factor cleanly, yielding rational roots without the quadratic formula. Given Data / Assumptions: Equation: 2x^2 − 11x + 15 = 0 Look for factorization into (2x − m)(x − n) with mn = 15 and 2n + m = 11. Concept / Approach:We seek integers m, n such that 2x^2 − 11x + 15 = (2x − 5)(x − 3). Then set each factor to zero to get the roots. This is quicker than the quadratic formula and equally valid. Step-by-Step Solution: 2x^2 − 11x + 15 = (2x − 5)(x − 3)Set factors to zero: 2x − 5 = 0 ⇒ x = 5/2; x − 3 = 0 ⇒ x = 3 Verification / Alternative check:Expand (2x − 5)(x − 3) = 2x^2 − 6x − 5x + 15 = 2x^2 − 11x + 15, confirming correctness. Why Other Options Are Wrong: Other sign variants do not satisfy the original equation when substituted. Common Pitfalls:Misassigning signs to factor pairs of 15, or mixing the coefficient 2 into the wrong factor, leading to cross-term errors. Final Answer:3 and 5/2

Question 6

Evaluate a^3 + 1/a^3 from a squared sum: If (a + 1/a)^2 = 3, find the value of a^3 + 1/a^3.

Accepted Answer

Introduction / Context:Power-sum identities allow computation of higher symmetric expressions from lower ones. Given (a + 1/a)^2, we can find (a + 1/a) and then use the identity for a^3 + 1/a^3 in terms of (a + 1/a). This avoids solving explicitly for a. Given Data / Assumptions: (a + 1/a)^2 = 3 a ≠ 0 Concept / Approach:Use two identities: (a + 1/a)^2 = a^2 + 2 + 1/a^2 ⇒ a^2 + 1/a^2 = 1. Also, a^3 + 1/a^3 = (a + 1/a)^3 − 3(a + 1/a). Knowing S = a + 1/a from S^2, we can compute the cube expression directly. Step-by-Step Solution: Let S = a + 1/a. Given S^2 = 3 ⇒ S = √3 or S = −√3a^3 + 1/a^3 = S^3 − 3SIf S = √3: S^3 − 3S = 3√3 − 3√3 = 0If S = −√3: (−3√3) − 3(−√3) = −3√3 + 3√3 = 0 Verification / Alternative check:From S^2 = 3 ⇒ a^2 + 1/a^2 = 1. Then (a + 1/a)(a^2 − 1 + 1/a^2) = a^3 + 1/a^3, which also evaluates to 0 when S = ±√3. Why Other Options Are Wrong: 10√3/3, 3√3, 6√3, 9: None align with the identity S^3 − 3S when S^2 = 3. Common Pitfalls:Assuming S = √3 only; even S = −√3 yields the same final value. Also, confusing a^3 + 1/a^3 with (a + 1/a)^3 directly, forgetting the −3(a + 1/a) term. Final Answer:0

Question 7

Homogeneous quadratic in x and y (ratio form): If 2x^2 − 7xy + 3y^2 = 0 for nonzero x and y, determine the possible values of the ratio x : y.

Accepted Answer

Introduction / Context:This problem asks for the ratio x : y that satisfies a homogeneous quadratic relation 2x^2 − 7xy + 3y^2 = 0. Because every term carries the same total degree, dividing through by y^2 (assuming y ≠ 0) reduces the equation to a quadratic in the single variable r = x / y. Solving that quadratic yields the admissible ratios, which must be presented as exact pairs x : y.Given Data / Assumptions: Equation: 2x^2 − 7xy + 3y^2 = 0. x and y are real and not both zero; take y ≠ 0 for ratio. We need all possible real ratios x : y that satisfy the equation. Concept / Approach:Introduce r = x / y. Divide the equation by y^2 to obtain a standard quadratic in r. Solve using the quadratic formula and convert each r back into the ratio x : y = r : 1 (and, if r is a fraction p/q, express as p : q).Step-by-Step Solution: Let r = x / y. Then 2r^2 − 7r + 3 = 0.Discriminant D = (−7)^2 − 4*2*3 = 49 − 24 = 25.r = [7 ± √25] / (4) = (7 ± 5) / 4 ⇒ r = 12/4 = 3 or r = 2/4 = 1/2.Thus x : y = 3 : 1 or x : y = 1 : 2. Verification / Alternative check:Substitute r = 3: 2*(9) − 7*(3) + 3 = 18 − 21 + 3 = 0. Substitute r = 1/2: 2*(1/4) − 7*(1/2) + 3 = 0.5 − 3.5 + 3 = 0. Both satisfy the equation exactly. Why Other Options Are Wrong: 3 : 2 or 2 : 3: These correspond to r = 3/2 or r = 2/3, neither satisfies 2r^2 − 7r + 3 = 0. 5 : 6: Gives r = 5/6, which does not solve the quadratic. Common Pitfalls:Forgetting to convert r values to simplified integer ratios, or losing a valid root by factoring incorrectly. Always check both quadratic roots and express as clean ratios. Final Answer: 3 : 1 and 1 : 2

Question 8

Condition for reciprocal roots (standard form): If the quadratic equation (x^2)/a + (x)/b + 1/c = 0 has one root equal to the reciprocal of the other (both nonzero), which condition among a, b, c must hold?

Accepted Answer

Introduction / Context:A quadratic px^2 + qx + r = 0 has roots that are reciprocals of each other (both nonzero) if and only if p = r. This follows from the Vieta relations: product of roots = r / p, which must equal 1 for reciprocal roots, hence r = p. We rewrite the given equation in standard form to identify p and r and then apply the criterion.Given Data / Assumptions: Equation: (x^2)/a + (x)/b + 1/c = 0. a, b, c ≠ 0; roots are finite and nonzero. We seek the relation among a, b, c ensuring reciprocal roots. Concept / Approach:Express in standard quadratic form px^2 + qx + r = 0. Here p = 1/a, q = 1/b, r = 1/c. For reciprocal roots, require r = p ⇒ 1/c = 1/a ⇒ a = c. Step-by-Step Solution: Identify coefficients: p = 1/a, q = 1/b, r = 1/c.Condition for reciprocal roots: r = p ⇒ 1/c = 1/a.Therefore, a = c. Verification / Alternative check:If a = c, then product of roots = r/p = (1/c) / (1/a) = a/c = 1, so roots are reciprocals (assuming both nonzero). The condition is both necessary and sufficient. Why Other Options Are Wrong: a = b or b = c: These do not force r = p. ac = 1: Irrelevant to r/p; it constrains magnitudes but not the required equality 1/c = 1/a. Common Pitfalls:Confusing the “sum of roots” condition with “reciprocal roots.” Only the product matters here. Ensure the quadratic is in standard form before applying criteria. Final Answer: a = c

Question 9

Find the other root (Vieta's use): One root of the quadratic equation x^2 − 5x + 6 = 0 is 3. Determine the other root.

Accepted Answer

Introduction / Context:The classic way to find the second root of a quadratic when one root is known is to use the sum and product of roots. For ax^2 + bx + c = 0 with roots r1 and r2, r1 + r2 = −b/a and r1*r2 = c/a. With r1 given, solve for r2 quickly without factoring from scratch.Given Data / Assumptions: Equation: x^2 − 5x + 6 = 0. One root r1 = 3. Concept / Approach:Use Vieta: Sum of roots = 5 (since −b/a = −(−5)/1 = 5). Then r2 = 5 − r1. Alternatively, factor the quadratic if convenient. Step-by-Step Solution: Sum of roots = 5 ⇒ r2 = 5 − r1 = 5 − 3 = 2.Check via product: r1*r2 = 3*2 = 6, which matches c/a = 6. Verification / Alternative check:Factorization: x^2 − 5x + 6 = (x − 2)(x − 3) = 0, confirming roots 2 and 3. Why Other Options Are Wrong: −2, 1, −1: Do not satisfy both sum 5 and product 6 with the given root 3. Common Pitfalls:Sign mistakes in using −b/a for the sum, or mixing up which coefficient is b. Always write the standard form and read off a, b, c carefully. Final Answer: 2

Question 10

Construct a quadratic from sum and product of roots: Form the quadratic equation whose roots have sum 6 and product −16.

Accepted Answer

Introduction / Context:Given the sum (S) and product (P) of roots, the monic quadratic is x^2 − Sx + P = 0. This is a direct application of Vieta’s relations and avoids computing the individual roots.Given Data / Assumptions: Sum S = 6. Product P = −16. Concept / Approach:Write x^2 − Sx + P = 0. Substitute S and P as provided, ensuring the signs are placed correctly (note that P is negative here). Step-by-Step Solution: General form: x^2 − Sx + P = 0.With S = 6 and P = −16: x^2 − 6x − 16 = 0. Verification / Alternative check:If roots are r1 and r2, then r1 + r2 = 6 and r1*r2 = −16. Expanding (x − r1)(x − r2) gives x^2 − (r1 + r2)x + r1r2 = x^2 − 6x − 16, as required. Why Other Options Are Wrong: x^2 + 6x − 16 = 0: Has sum −6. x^2 − √3x − 16 = 0: Introduces an irrelevant irrational coefficient. None of these: Incorrect because x^2 − 6x − 16 = 0 matches exactly. Common Pitfalls:Mixing the sign of P or writing x^2 − Sx − P by mistake. Always align with x^2 − Sx + P. Final Answer: x^2 − 6x − 16 = 0

Question 11

Compare x and y (define the answer codes): I. 4x + 7y = 209 II. 12x − 14y = −38 Choose one option: A) x > y  B) x < y  C) x = y  D) Relationship cannot be determined

Accepted Answer

Introduction / Context:We are given two linear equations in x and y and asked to compare their values. The most reliable approach is to solve the system exactly and then compare the numerical values. The answer codes are provided explicitly to avoid ambiguity common in comparison problems.Given Data / Assumptions: I: 4x + 7y = 209 II: 12x − 14y = −38 Real numbers x, y. Concept / Approach:Use elimination or substitution to solve the simultaneous linear equations. Then compare the found x and y to select the appropriate code (A, B, C, or D). Step-by-Step Solution: Multiply I by 3: 12x + 21y = 627.Subtract II: (12x + 21y) − (12x − 14y) = 627 − (−38) ⇒ 35y = 665 ⇒ y = 19.Back-substitute in I: 4x + 7*19 = 209 ⇒ 4x + 133 = 209 ⇒ 4x = 76 ⇒ x = 19.Therefore x = y = 19. Verification / Alternative check:Check in II: 12*19 − 14*19 = (12 − 14)*19 = −2*19 = −38, which matches II exactly. Why Other Options Are Wrong: x > y or x < y: Both false since x and y are equal. Relationship cannot be determined: The system gives a unique solution, so the relationship is determined. Common Pitfalls:Arithmetic slips during elimination (especially sign errors when subtracting). Always verify by substituting the result into both equations. Final Answer: x = y

Question 12

Compare x and y (define the answer codes): I. 18x^2 + 18x + 4 = 0 II. 12y^2 + 29y + 14 = 0 x and y are real roots (any root from each). Choose: A) x > y  B) x < y  C) x = y  D) Relationship cannot be determined

Accepted Answer

Introduction / Context:This comparison asks us to pick one root from each quadratic and compare them. Since each quadratic has two distinct real roots, the relationship may depend on which roots are chosen unless a strict ordering holds across all combinations. We must compute the roots or characterize their ranges.Given Data / Assumptions: I: 18x^2 + 18x + 4 = 0 ⇒ 9x^2 + 9x + 2 = 0. II: 12y^2 + 29y + 14 = 0. Any real root x from I and any real root y from II can be selected. Concept / Approach:Find both roots for each quadratic using the quadratic formula and list their numerical values. If there is overlap (equality) or changing order depending on choices, the relationship cannot be uniquely determined. Step-by-Step Solution: For I: 9x^2 + 9x + 2 = 0 ⇒ Δ = 81 − 72 = 9.x = [−9 ± 3]/(18) ⇒ x ∈ {−2/3, −1/3} ≈ {−0.6667, −0.3333}.For II: 12y^2 + 29y + 14 = 0 ⇒ Δ = 29^2 − 4*12*14 = 169.y = [−29 ± 13]/(24) ⇒ y ∈ {−7/4, −2/3} ≈ {−1.75, −0.6667}.Observe: x can be −2/3 or −1/3; y can be −7/4 or −2/3. Depending on choices, x can be greater, equal, or less (e.g., x = −2/3 equals y = −2/3; x = −1/3 > y = −2/3; but compared to y = −1.75, both x values are greater). No single ordering fits all selections. Verification / Alternative check:Tabulate pairs to see the variability. The presence of equality (−2/3 equals −2/3) already prevents a strict inequality conclusion. Why Other Options Are Wrong: x > y, x < y, x = y: Each can occur for specific choices, but none holds universally across all root pairings. Common Pitfalls:Assuming a single root per equation or overlooking that either root may be picked. For comparison questions, confirm if the order is consistent for all combinations. Final Answer: Relationship cannot be determined

Question 13

Compare x and y (define the answer codes): I. 8x^2 + 6x = 5 II. 12y^2 − 22y + 8 = 0 x and y are real roots (any root from each). Choose: A) x > y  B) x < y  C) x = y  D) Relationship cannot be determined

Accepted Answer

Introduction / Context:Each quadratic yields two real roots. Unless one interval of roots lies entirely above or below the other, the pairwise comparison will depend on which roots are chosen. We compute exact roots to check for consistent ordering.Given Data / Assumptions: I: 8x^2 + 6x − 5 = 0. II: 12y^2 − 22y + 8 = 0. Any real root from each equation may be chosen independently. Concept / Approach:Apply the quadratic formula to both equations and list all roots. Examine whether a single inequality holds across every combination of choices; if not, the relationship is indeterminate. Step-by-Step Solution: I: Δ = 6^2 − 4*8*(−5) = 36 + 160 = 196 ⇒ √Δ = 14.x = [−6 ± 14]/(16) ⇒ x ∈ {0.5, −1.25}.II: Δ = (−22)^2 − 4*12*8 = 484 − 384 = 100 ⇒ √Δ = 10.y = [22 ± 10]/24 ⇒ y ∈ {4/3 ≈ 1.3333, 1/2 = 0.5}.Comparisons vary: x = 0.5 equals y = 0.5; x = −1.25 < y for both y values; x = 0.5 < 4/3. Therefore, a single universal relation cannot be asserted. Verification / Alternative check:Construct a 2×2 table of choices confirming equality in one pair and strict inequalities in others with different directions. Why Other Options Are Wrong: x > y, x < y, x = y: Each is true for some particular pairing, but not for all pairings. Common Pitfalls:Assuming the principal (greater) root is always chosen; the problem statement allows any root, making a single comparison impossible. Final Answer: Relationship cannot be determined

Question 14

Compare x and y (define the answer codes): I. 16x^2 + 20x + 6 = 0 II. 10y^2 + 38y + 24 = 0 x and y are real roots (any root from each). Choose: A) x > y  B) x < y  C) x = y  D) Relationship cannot be determined

Accepted Answer

Introduction / Context:We compare any root x from the first quadratic with any root y from the second. If every x exceeds every y (or vice versa), then the relationship is determined. Compute all roots and compare the ranges systematically.Given Data / Assumptions: I: 16x^2 + 20x + 6 = 0. II: 10y^2 + 38y + 24 = 0. Real roots only; any choice per equation. Concept / Approach:Find both roots of each quadratic precisely. Determine the interval of possible x values and the interval of possible y values. If the minimum x is greater than the maximum y, then x > y for all combinations. Step-by-Step Solution: I: Divide by 2 ⇒ 8x^2 + 10x + 3 = 0. Δ = 10^2 − 4*8*3 = 100 − 96 = 4 ⇒ √Δ = 2.x = [−10 ± 2]/(16) ⇒ x ∈ {−12/16 = −0.75, −8/16 = −0.5}.II: Divide by 2 ⇒ 5y^2 + 19y + 12 = 0. Δ = 19^2 − 4*5*12 = 361 − 240 = 121 ⇒ √Δ = 11.y = [−19 ± 11]/(10) ⇒ y ∈ {−30/10 = −3, −8/10 = −0.8}.Compare ranges: min(x) = −0.75, max(x) = −0.5; min(y) = −3, max(y) = −0.8. Since −0.75 > −0.8 and −0.5 > −0.8, every x is greater than every y. Verification / Alternative check:Test all combinations explicitly: {−0.75, −0.5} minus {−3, −0.8} ⇒ all differences are positive. Why Other Options Are Wrong: x < y or x = y: Contradicted by computed values. Relationship cannot be determined: Here it is determined since intervals do not overlap. Common Pitfalls:Arithmetic errors in discriminants or sign mistakes when dividing. Always simplify first to reduce errors. Final Answer: x > y

Question 15

Compare x and y (define the answer codes): I. 17x^2 + 48x = 9 II. 13y^2 = 32y − 12 Choose: A) x > y  B) x < y  C) x = y  D) Relationship cannot be determined

Accepted Answer

Introduction / Context:We compare solutions from two quadratic relations. If every possible x is strictly less than every possible y, we can conclude x < y. Compute the roots for both to compare their ranges.Given Data / Assumptions: I: 17x^2 + 48x − 9 = 0. II: 13y^2 − 32y + 12 = 0. Concept / Approach:Use quadratic formula to get all roots. Compare the largest x to the smallest y; if the largest x is still less than the smallest y, then every x is less than every y. Step-by-Step Solution: I: Δ = 48^2 − 4*17*(−9) = 2304 + 612 = 2916 ⇒ √Δ = 54.x = [−48 ± 54]/(34) ⇒ x ∈ { (6/34) ≈ 0.17647, (−102/34) = −3 }.II: Δ = (−32)^2 − 4*13*12 = 1024 − 624 = 400 ⇒ √Δ = 20.y = [32 ± 20]/(26) ⇒ y ∈ { 52/26 = 2, 12/26 ≈ 0.46154 }.Largest x is ≈ 0.17647; smallest y is ≈ 0.46154. Therefore, even the largest x is less than the smallest y ⇒ x < y for all choices. Verification / Alternative check:Check all pairings: {−3, 0.17647} versus {0.46154, 2}; all comparisons show x < y. Why Other Options Are Wrong: x > y or x = y: Contradicted by numeric ranges. Relationship cannot be determined: Not the case; strict separation exists. Common Pitfalls:Not identifying the smallest y and largest x correctly; mixing signs or miscomputing discriminants can flip conclusions. Final Answer: x < y

Question 16

Roots of a parameterized quadratic: Find the roots of a^2 x^2 − 3ab x + 2b^2 = 0 in terms of a and b (assume a ≠ 0).

Accepted Answer

Introduction / Context:We are to solve a^2 x^2 − 3ab x + 2b^2 = 0 for x in terms of parameters a and b. Treat this as a standard quadratic with respect to x and apply the quadratic formula. Parameterized quadratics commonly appear in algebraic identities and comparisons.Given Data / Assumptions: Quadratic: a^2 x^2 − 3ab x + 2b^2 = 0. a ≠ 0 (to ensure a^2 is invertible); b can be any real. Concept / Approach:Identify coefficients: A = a^2, B = −3ab, C = 2b^2. Use x = [−B ± √(B^2 − 4AC)] / (2A) and simplify by factoring ab where possible. Step-by-Step Solution: Discriminant: B^2 − 4AC = (−3ab)^2 − 4*a^2*2b^2 = 9a^2b^2 − 8a^2b^2 = a^2b^2.√Discriminant = |ab|; algebraically we take ±ab in the formula.x = [3ab ± ab]/(2a^2) = (ab(3 ± 1))/(2a^2) = (b/a) * (3 ± 1)/2.Thus roots: x = 2b/a and x = b/a. Verification / Alternative check:Plug x = 2b/a: a^2*(4b^2/a^2) − 3ab*(2b/a) + 2b^2 = 4b^2 − 6b^2 + 2b^2 = 0. Similarly for x = b/a: a^2*(b^2/a^2) − 3ab*(b/a) + 2b^2 = b^2 − 3b^2 + 2b^2 = 0. Why Other Options Are Wrong: Signs with negatives do not satisfy the original quadratic. “None of these” is incorrect since the explicit pair 2b/a and b/a works. Common Pitfalls:Dropping parameters during simplification or mishandling the square root of a^2b^2. Keep factors symbolic and simplify carefully. Final Answer: 2b/a , b/a

Question 17

Compute α^3 + β^3 from ax^2 + bx + c = 0: If α and β are roots of ax^2 + bx + c = 0 (a ≠ 0), find α^3 + β^3 in terms of a, b, c.

Accepted Answer

Introduction / Context:Power sums of roots can be expressed using elementary symmetric sums. For α, β as roots of ax^2 + bx + c = 0, we know α + β = −b/a and αβ = c/a. Using the identity α^3 + β^3 = (α + β)^3 − 3αβ(α + β), we can express the result entirely in a, b, c.Given Data / Assumptions: α + β = −b/a. αβ = c/a. a ≠ 0. Concept / Approach:Apply the cube sum identity and substitute Vieta’s formulas carefully. Simplify to a single rational expression with denominator a^3. Step-by-Step Solution: α^3 + β^3 = (α + β)^3 − 3αβ(α + β).= (−b/a)^3 − 3*(c/a)*(−b/a) = −b^3/a^3 + 3bc/a^2.Put over a^3: (−b^3 + 3abc)/a^3 = b(3ac − b^2)/a^3. Verification / Alternative check:Test with a simple quadratic, e.g., x^2 − sx + p = 0 (a = 1, b = −s, c = p). Then α^3 + β^3 = s^3 − 3ps, which matches b(3ac − b^2)/a^3 after substituting. Why Other Options Are Wrong: b(b^2 − 3ac)/a^3: Sign reversed. b(3ac + b^2)/a^3: Adds instead of subtracts, not supported by identity. None of these: Incorrect; the expression b(3ac − b^2)/a^3 is exact. Common Pitfalls:Sign mistakes when cubing (−b/a) and distributing the negative in −3αβ(α + β). Keep denominators consistent. Final Answer: b(3ac − b^2)/a^3

Question 18

Real factorization condition: For which values of k does the quadratic polynomial 3z^2 + 5z + k factor over the reals (i.e., has real linear factors)?

Accepted Answer

Introduction / Context:A quadratic az^2 + bz + c factors into real linear factors iff its discriminant Δ = b^2 − 4ac is nonnegative. This is the necessary and sufficient condition for real roots (counting multiplicity). Apply it directly to 3z^2 + 5z + k.Given Data / Assumptions: a = 3, b = 5, c = k. Real coefficients; want real roots. Concept / Approach:Impose b^2 − 4ac ≥ 0 ⇒ 25 − 4*3*k ≥ 0 ⇒ 25 − 12k ≥ 0 ⇒ k ≤ 25/12. Step-by-Step Solution: Δ = 5^2 − 4*3*k = 25 − 12k.Require Δ ≥ 0 ⇒ 25 − 12k ≥ 0 ⇒ 12k ≤ 25 ⇒ k ≤ 25/12. Verification / Alternative check:At k = 25/12, Δ = 0, giving a repeated real root; for k less than this value, Δ > 0 gives two distinct real roots. For k > 25/12, Δ < 0, so no real factorization. Why Other Options Are Wrong: k ≤ 25/6 or k ≥ 25/6: Wrong threshold; uses 4ac with a mistaken coefficient. k ≥ 25/12: Reverses the inequality; would imply nonreal roots for most values. Common Pitfalls:Arithmetic errors in computing 4ac or reversing the inequality sign when moving terms. Keep Δ ≥ 0 straight. Final Answer: k ≤ 25/12

Question 19

Equal roots condition: The quadratic x^2 + px + q = 0 has equal roots if and only if:

Accepted Answer

Introduction / Context:A quadratic has equal (repeated) roots when its discriminant is zero. For x^2 + px + q = 0, the discriminant is Δ = p^2 − 4q. Setting Δ = 0 gives the precise condition connecting p and q.Given Data / Assumptions: Quadratic: x^2 + px + q = 0. Real coefficients; equal roots desired. Concept / Approach:Use Δ = p^2 − 4q = 0 ⇒ p^2 = 4q. This is necessary and sufficient. Step-by-Step Solution: Compute Δ: p^2 − 4q.Set Δ = 0 ⇒ p^2 = 4q. Verification / Alternative check:If p^2 = 4q, then the quadratic is (x + p/2)^2 = 0, showing a double root at x = −p/2. Why Other Options Are Wrong: p^2 = 2q or p^2 = −2q or p^2 = −4q: Do not make Δ = 0; they lead to unequal or complex roots depending on signs. Common Pitfalls:Confusing the sign in Δ or misplacing the 4. The standard discriminant for ax^2 + bx + c is b^2 − 4ac. Final Answer: p^2 = 4q

Question 20

Consecutive positive odd integers (sum of squares = 290): Find two consecutive positive odd integers whose squares add up to 290.

Accepted Answer

Introduction / Context:Let the odd integers be n and n + 2. Translating the statement “their squares add to 290” yields a quadratic equation in n. Solving it produces the valid positive odd integers. This is a standard modeling step from words to algebra.Given Data / Assumptions: Integers are consecutive and odd: n and n + 2. n > 0. n^2 + (n + 2)^2 = 290. Concept / Approach:Expand, simplify to a quadratic in n, and solve. Discard negative solutions since the integers must be positive. Step-by-Step Solution: n^2 + (n + 2)^2 = 290 ⇒ n^2 + n^2 + 4n + 4 = 290.2n^2 + 4n − 286 = 0 ⇒ divide by 2 ⇒ n^2 + 2n − 143 = 0.Discriminant D = 2^2 + 4*143 = 4 + 572 = 576 ⇒ √D = 24.n = [−2 ± 24]/2 ⇒ n = 11 or n = −13 (reject the negative).Thus the integers are 11 and 13. Verification / Alternative check:11^2 + 13^2 = 121 + 169 = 290 as required. Why Other Options Are Wrong: 13, 15: 169 + 225 = 394, not 290. 9, 11: 81 + 121 = 202, not 290. None of these: Incorrect since 11 and 13 satisfy the condition. Common Pitfalls:Forgetting that consecutive odds differ by 2 (not 1), or arithmetic mistakes while expanding squares. Final Answer: 11, 13

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