Q: Comparison of x and y from two quadratics (use mapping below): I. 16x^2 + 20x + 6 = 0 II. 10y^2 + 38y + 24 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context: The task is to compare the possible values of x and y when each variable must satisfy its own quadratic. Because a quadratic can have two roots, we must check all admissible x and y values and see if a single consistent relation holds in every case. Given Data / Assumptions: I: 16x^2 + 20x + 6 = 0 II: 10y^2 + 38y + 24 = 0 Concept / Approach: Factor (or reduce) each quadratic to find its real roots. Then compare every x-root with every y-root. If for all combinations x is greater than y, we can conclude x > y unequivocally. Step-by-Step Solution: I: Divide by 2 → 8x^2 + 10x + 3 = 0 = (4x + 3)(2x + 1)So x ∈ {−3/4 = −0.75, −1/2 = −0.5}II: Divide by 2 → 5y^2 + 19y + 12 = 0 = (5y + 4)(y + 3)So y ∈ {−4/5 = −0.8, −3}Compare: each x (−0.75, −0.5) is greater than each y (−0.8, −3). Verification / Alternative check: Numerically: max(y) = −0.8, min(x) = −0.75; since −0.75 > −0.8, even the least x exceeds the greatest y. Hence x > y for all admissible cases. Why Other Options Are Wrong: “x y

Q: Comparison of x and y from two quadratics (use mapping below): I. 18x^2 + 18x + 4 = 0 II. 12y^2 + 29y + 14 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context: When both x and y are defined by quadratics with two real roots, comparisons must consider all combinations. If some combinations yield equality and others yield strict inequality, the overall relationship is indeterminate. Given Data / Assumptions: I: 18x^2 + 18x + 4 = 0 II: 12y^2 + 29y + 14 = 0 Concept / Approach: Find the roots of each. Then examine whether a single relation (>, y. Different relations occur. Verification / Alternative check: Narrowest gap occurs at the common root −2/3. Since equality is possible and x > y also occurs, no single relation always holds. Why Other Options Are Wrong: Choosing x > y or x = y ignores that both cases occur; x < y never occurs for any admissible pair. Common Pitfalls: Concluding based on a single pair, missing the shared root case that forces equality. Final Answer: Relationship cannot be determined

Q: Solve the linear system and compare x and y (use mapping below): I. 4x + 7y = 209 II. 12x − 14y = −38 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context: For two linear equations in two variables, there is a unique solution when the lines are non-parallel. Once x and y are found, the comparison is direct and definitive. Given Data / Assumptions: 4x + 7y = 209 12x − 14y = −38 Concept / Approach: Use elimination by aligning coefficients of y. Add the equations to eliminate y, compute x, then back-substitute to find y. Finally, compare x and y numerically. Step-by-Step Solution: From II: 12x − 14y = −38 ⇒ divide by 2 ⇒ 6x − 7y = −19Add with I: (4x + 7y) + (6x − 7y) = 209 + (−19) ⇒ 10x = 190x = 19Back-substitute: 4*19 + 7y = 209 ⇒ 76 + 7y = 209 ⇒ 7y = 133 ⇒ y = 19Hence x = y Verification / Alternative check: Plug x = y = 19 into both equations to confirm: 76 + 133 = 209 and 228 − 266 = −38. Both hold. Why Other Options Are Wrong: Since x and y are exactly equal, “greater than”, “less than”, or “indeterminate” are incorrect. Common Pitfalls: Arithmetic slips when dividing or adding equations. Keep track of signs carefully. Final Answer: x = y

Q: Comparison of x and y from two equations (use mapping below): I. 17x^2 + 48x = 9 II. 13y^2 = 32y − 12 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context: Each variable is governed by a quadratic; we must list all possible roots for x and y, then check whether a single inequality relation holds for every admissible pair (x, y). Given Data / Assumptions: I: 17x^2 + 48x − 9 = 0 II: 13y^2 − 32y + 12 = 0 Concept / Approach: Compute discriminants to find exact roots. Then compare the sets: if every x is less than every y, conclude x < y conclusively. Step-by-Step Solution: I: D = 48^2 − 4*17*(−9) = 2304 + 612 = 2916 = 54^2x = [−48 ± 54]/(2*17) ⇒ x ∈ {3/17 ≈ 0.1765, −3}II: D = (−32)^2 − 4*13*12 = 1024 − 624 = 400 = 20^2y = [32 ± 20]/(2*13) ⇒ y ∈ {2, 6/13 ≈ 0.4615}Compare every pair: −3 y” or “x = y” never occurs with these root sets; “indeterminate” is wrong because the inequality is consistent. Common Pitfalls: Missing one root or miscomputing the discriminant can flip the conclusion; compute precisely. Final Answer: x < y

Q: Comparison of x and y from two equations (use mapping below): I. 8x^2 + 6x = 5 II. 12y^2 − 22y + 8 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context: When multiple roots exist for both variables, the relation between x and y may vary across admissible pairs. We conclude “indeterminate” if both equality and strict inequality occur among the valid combinations. Given Data / Assumptions: I: 8x^2 + 6x − 5 = 0 II: 12y^2 − 22y + 8 = 0 Concept / Approach: Find roots of both equations, then compare. A single consistent relation is needed to declare x > y, x < y, or x = y; otherwise choose “Relationship cannot be determined.” Step-by-Step Solution: I: D = 6^2 + 4*8*5 = 36 + 160 = 196 = 14^2x = [−6 ± 14]/16 ⇒ x ∈ {0.5, −1.25}II: Divide by 2 ⇒ 6y^2 − 11y + 4 = 0, D = 121 − 96 = 25 = 5^2y = [11 ± 5]/12 ⇒ y ∈ {4/3 ≈ 1.333, 1/2 = 0.5}Comparisons: x = 0.5 with y = 0.5 gives equality; x = −1.25 with y = 0.5 gives x y ignores actual values; choosing x = y ignores the other x-root; choosing x < y ignores the equality case. Common Pitfalls: Overlooking repeated roots or failing to consider every possible pair when deciding the relation. Final Answer: Relationship cannot be determined

Question 1

Exactly one real root (repeated) condition: Determine all real values of p such that x^2 + 5px + 16 = 0 has exactly one real root (i.e., equal roots).

Accepted Answer

Introduction / Context: “Exactly one real root” for a quadratic means the graph touches the x-axis at a single point: the discriminant must be zero. We find all p that make Δ = 0 for x^2 + 5px + 16 = 0. Given Data / Assumptions: a = 1, b = 5p, c = 16. p is real. Concept / Approach: Set Δ = b^2 − 4ac = 0 and solve for p. This yields the exact parameter values where the quadratic has a repeated real root. Step-by-Step Solution: Δ = (5p)^2 − 4*1*16 = 25p^2 − 64.Set Δ = 0 ⇒ 25p^2 − 64 = 0 ⇒ 25p^2 = 64 ⇒ p^2 = 64/25.Therefore p = ± 8/5. Verification / Alternative check: For p = 8/5, the quadratic is x^2 + 8x + 16 = 0 ⇒ (x + 4)^2 = 0. For p = −8/5, x^2 − 8x + 16 = 0 ⇒ (x − 4)^2 = 0. Both have a single (repeated) real root. Why Other Options Are Wrong: −8/5 < p < 8/5: Gives Δ < 0 (no real roots). p ≤ −8/5 or p ≥ 8/5: Includes Δ > 0 as well as Δ = 0; not “exactly one.” No real value of p: False; two real values exist. Common Pitfalls: Confusing “at least one real root” (Δ ≥ 0) with “exactly one real root” (Δ = 0). Here the equality is essential. Final Answer: p = 8/5 or p = −8/5

Question 2

Quadratic roots identity (Vieta's relations): If α and β are the roots of ax^2 + bx + c = 0 (a ≠ 0), find the value of (α/β) + (β/α) in terms of a, b, c.

Accepted Answer

Introduction / Context: This problem checks your comfort with Vieta's relations for quadratics and your ability to manipulate symmetric expressions in the roots. Instead of solving for the roots explicitly, we will express (α/β) + (β/α) directly using a, b, and c. Given Data / Assumptions: Quadratic: ax^2 + bx + c = 0 with a ≠ 0. Roots: α and β (possibly real or complex). We require (α/β) + (β/α). Concept / Approach: By Vieta's relations, α + β = −b/a and αβ = c/a. Note that (α/β) + (β/α) = (α^2 + β^2)/(αβ). Also, α^2 + β^2 = (α + β)^2 − 2αβ. Substitute Vieta's expressions and simplify carefully to avoid sign errors. Step-by-Step Solution: α + β = −b/aαβ = c/aα^2 + β^2 = (α + β)^2 − 2αβ = (b^2/a^2) − 2(c/a)(α/β) + (β/α) = (α^2 + β^2)/(αβ) = [ (b^2/a^2) − 2(c/a) ] / (c/a)= (b^2/a − 2c)/c = (b^2 − 2ac)/(ac) Verification / Alternative check: Plugging a simple example (e.g., a = 1, b = −3, c = 1 with roots 1 and 2) confirms the formula numerically equals (1/2 + 2/1) = 2.5 and (b^2 − 2ac)/(ac) = (9 − 2)/1 = 7; wait, with a = 1, b = −3, c = 2 (roots 1 and 2), (α/β + β/α) = 1/2 + 2 = 2.5 and (b^2 − 2ac)/(ac) = (9 − 4)/(2) = 2.5—consistent. Why Other Options Are Wrong: Dividing by 2ac or adding 2ac in the numerator changes the value; these arise from algebra slips in the transformation from α^2 + β^2 to (α/β + β/α). Common Pitfalls: Forgetting that α^2 + β^2 = (α + β)^2 − 2αβ, or mishandling complex fractions while clearing denominators. Final Answer: (b^2 - 2ac) / (ac)

Question 3

Construct the quadratic equation whose roots are √2 and 2√2. Express the equation with integer constant term and a rational middle coefficient if possible.

Accepted Answer

Introduction / Context: Building a quadratic from its roots is a classic application of Vieta's relations. If r1 and r2 are the roots, then the monic quadratic is x^2 − (r1 + r2)x + r1 r2 = 0. The task uses irrational roots √2 and 2√2, but the product becomes an integer, simplifying the constant term. Given Data / Assumptions: Roots: √2 and 2√2. Monic quadratic form: x^2 − (sum)x + product = 0. Concept / Approach: Compute the sum and product of the given roots, and substitute into the monic quadratic template. Double-check arithmetic with radicals to avoid mistakes. Step-by-Step Solution: Sum = √2 + 2√2 = 3√2Product = √2 * 2√2 = 2*(√2*√2) = 2*2 = 4Quadratic: x^2 − (3√2)x + 4 = 0 Verification / Alternative check: Substitute x = √2: LHS = 2 − 3√2*√2 + 4 = 2 − 6 + 4 = 0. Substitute x = 2√2: LHS = 8 − 3√2*2√2 + 4 = 8 − 12 + 4 = 0. Both roots satisfy the equation. Why Other Options Are Wrong: Sign errors on the middle term or constant produce equations that do not vanish at both roots; testing either root quickly reveals nonzero residuals. Common Pitfalls: Miscomputing the product of radicals (√2 * 2√2 = 4, not 2√4). Keep the rule √2*√2 = 2 in mind. Final Answer: x^2 − 3√2 x + 4 = 0

Question 4

Comparison of x and y from two quadratics (use mapping below): I. 16x^2 + 20x + 6 = 0 II. 10y^2 + 38y + 24 = 0  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context: The task is to compare the possible values of x and y when each variable must satisfy its own quadratic. Because a quadratic can have two roots, we must check all admissible x and y values and see if a single consistent relation holds in every case. Given Data / Assumptions: I: 16x^2 + 20x + 6 = 0 II: 10y^2 + 38y + 24 = 0 Concept / Approach: Factor (or reduce) each quadratic to find its real roots. Then compare every x-root with every y-root. If for all combinations x is greater than y, we can conclude x > y unequivocally. Step-by-Step Solution: I: Divide by 2 → 8x^2 + 10x + 3 = 0 = (4x + 3)(2x + 1)So x ∈ {−3/4 = −0.75, −1/2 = −0.5}II: Divide by 2 → 5y^2 + 19y + 12 = 0 = (5y + 4)(y + 3)So y ∈ {−4/5 = −0.8, −3}Compare: each x (−0.75, −0.5) is greater than each y (−0.8, −3). Verification / Alternative check: Numerically: max(y) = −0.8, min(x) = −0.75; since −0.75 > −0.8, even the least x exceeds the greatest y. Hence x > y for all admissible cases. Why Other Options Are Wrong: “x < y” and “x = y” are contradicted by direct comparison; “Relationship cannot be determined” is invalid because the relationship is consistent across all root pairs. Common Pitfalls: Stopping after one comparison. You must ensure the relation holds for every root combination. Final Answer: x > y

Question 5

Comparison of x and y from two quadratics (use mapping below): I. 18x^2 + 18x + 4 = 0 II. 12y^2 + 29y + 14 = 0  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context: When both x and y are defined by quadratics with two real roots, comparisons must consider all combinations. If some combinations yield equality and others yield strict inequality, the overall relationship is indeterminate. Given Data / Assumptions: I: 18x^2 + 18x + 4 = 0 II: 12y^2 + 29y + 14 = 0 Concept / Approach: Find the roots of each. Then examine whether a single relation (>, <, or =) holds for every admissible pair (x, y). If not, choose “Relationship cannot be determined.” Step-by-Step Solution: I: Divide by 2 → 9x^2 + 9x + 2 = 0 = (3x + 1)(3x + 2)x ∈ {−1/3, −2/3}II: Discriminant = 29^2 − 4*12*14 = 169 ⇒ √D = 13y = (−29 ± 13)/24 ⇒ y ∈ {−2/3, −7/4}Comparisons: x = −2/3 and y = −2/3 give equality; x = −1/3 compared with y = −2/3 gives x > y. Different relations occur. Verification / Alternative check: Narrowest gap occurs at the common root −2/3. Since equality is possible and x > y also occurs, no single relation always holds. Why Other Options Are Wrong: Choosing x > y or x = y ignores that both cases occur; x < y never occurs for any admissible pair. Common Pitfalls: Concluding based on a single pair, missing the shared root case that forces equality. Final Answer: Relationship cannot be determined

Question 6

Solve the linear system and compare x and y (use mapping below): I. 4x + 7y = 209 II. 12x − 14y = −38  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context: For two linear equations in two variables, there is a unique solution when the lines are non-parallel. Once x and y are found, the comparison is direct and definitive. Given Data / Assumptions: 4x + 7y = 209 12x − 14y = −38 Concept / Approach: Use elimination by aligning coefficients of y. Add the equations to eliminate y, compute x, then back-substitute to find y. Finally, compare x and y numerically. Step-by-Step Solution: From II: 12x − 14y = −38 ⇒ divide by 2 ⇒ 6x − 7y = −19Add with I: (4x + 7y) + (6x − 7y) = 209 + (−19) ⇒ 10x = 190x = 19Back-substitute: 4*19 + 7y = 209 ⇒ 76 + 7y = 209 ⇒ 7y = 133 ⇒ y = 19Hence x = y Verification / Alternative check: Plug x = y = 19 into both equations to confirm: 76 + 133 = 209 and 228 − 266 = −38. Both hold. Why Other Options Are Wrong: Since x and y are exactly equal, “greater than”, “less than”, or “indeterminate” are incorrect. Common Pitfalls: Arithmetic slips when dividing or adding equations. Keep track of signs carefully. Final Answer: x = y

Question 7

Comparison of x and y from two equations (use mapping below): I. 17x^2 + 48x = 9 II. 13y^2 = 32y − 12  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context: Each variable is governed by a quadratic; we must list all possible roots for x and y, then check whether a single inequality relation holds for every admissible pair (x, y). Given Data / Assumptions: I: 17x^2 + 48x − 9 = 0 II: 13y^2 − 32y + 12 = 0 Concept / Approach: Compute discriminants to find exact roots. Then compare the sets: if every x is less than every y, conclude x < y conclusively. Step-by-Step Solution: I: D = 48^2 − 4*17*(−9) = 2304 + 612 = 2916 = 54^2x = [−48 ± 54]/(2*17) ⇒ x ∈ {3/17 ≈ 0.1765, −3}II: D = (−32)^2 − 4*13*12 = 1024 − 624 = 400 = 20^2y = [32 ± 20]/(2*13) ⇒ y ∈ {2, 6/13 ≈ 0.4615}Compare every pair: −3 < 0.4615 and −3 < 2; 0.1765 < 0.4615 and 0.1765 < 2. Verification / Alternative check: Note the largest x is about 0.1765 while the smallest y is about 0.4615, ensuring x < y for all combinations. Why Other Options Are Wrong: “x > y” or “x = y” never occurs with these root sets; “indeterminate” is wrong because the inequality is consistent. Common Pitfalls: Missing one root or miscomputing the discriminant can flip the conclusion; compute precisely. Final Answer: x < y

Question 8

Comparison of x and y from two equations (use mapping below): I. 8x^2 + 6x = 5 II. 12y^2 − 22y + 8 = 0  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context: When multiple roots exist for both variables, the relation between x and y may vary across admissible pairs. We conclude “indeterminate” if both equality and strict inequality occur among the valid combinations. Given Data / Assumptions: I: 8x^2 + 6x − 5 = 0 II: 12y^2 − 22y + 8 = 0 Concept / Approach: Find roots of both equations, then compare. A single consistent relation is needed to declare x > y, x < y, or x = y; otherwise choose “Relationship cannot be determined.” Step-by-Step Solution: I: D = 6^2 + 4*8*5 = 36 + 160 = 196 = 14^2x = [−6 ± 14]/16 ⇒ x ∈ {0.5, −1.25}II: Divide by 2 ⇒ 6y^2 − 11y + 4 = 0, D = 121 − 96 = 25 = 5^2y = [11 ± 5]/12 ⇒ y ∈ {4/3 ≈ 1.333, 1/2 = 0.5}Comparisons: x = 0.5 with y = 0.5 gives equality; x = −1.25 with y = 0.5 gives x < y. Verification / Alternative check: Since both equality and strict inequality (x < y) occur, no single relation always holds across all valid pairs. Why Other Options Are Wrong: Choosing x > y ignores actual values; choosing x = y ignores the other x-root; choosing x < y ignores the equality case. Common Pitfalls: Overlooking repeated roots or failing to consider every possible pair when deciding the relation. Final Answer: Relationship cannot be determined

Quadratic Equation Questions