Q: Compare x and y (define the answer codes): I. x^2 − 4 = 0 II. y^2 + 6y + 9 = 0 Roots are real (any root from each). Choose: A) x > y B) x < y C) x = y D) Relationship cannot be determined

Introduction / Context: The first quadratic yields two symmetric roots ±2, and the second is a perfect square with a repeated root at −3. We must compare any root x from I with any root y from II and select the correct relation that always holds. Given Data / Assumptions: I: x^2 − 4 = 0 ⇒ x = 2 or x = −2. II: y^2 + 6y + 9 = 0 ⇒ (y + 3)^2 = 0 ⇒ y = −3. Concept / Approach: List all possible x values and the (single) y value. Compare systematically. If the smallest possible x is still greater than y, then x > y regardless of choice. Step-by-Step Solution: Possible x: {−2, 2}; y = −3.Compare: −2 > −3 and 2 > −3.Thus, for any choice, x > y. Verification / Alternative check: Draw a number line: y fixed at −3; both x values lie to the right (−2 and 2), confirming x > y. Why Other Options Are Wrong: x y

Q: Comparison of x and y from two statements (use the mapping below): I. x^2 = 729 II. y = √729 (principal square root) Use this mapping for the options: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined from the information g…

Introduction / Context: This comparison problem provides two separate statements about x and y. You must deduce the relationship between x and y without assuming anything beyond standard definitions. The twist is that I gives a squared equation with two solutions, while II fixes y via a principal square root, which is always the non-negative value. Given Data / Assumptions: I: x^2 = 729 II: y = √729, the principal square root Principal square root √n is defined as the unique non-negative number whose square is n. Concept / Approach: From I, solve for x by taking square roots, remembering both positive and negative roots. From II, determine y directly using the principal root. Then test all possible values of x against y to see if a single strict relation (>, <, =) always holds. If different cases give different relations, the comparison is indeterminate. Step-by-Step Solution: From I: x^2 = 729 ⇒ x = ±27From II: y = √729 = 27 (principal root)Case 1: x = 27 ⇒ x = yCase 2: x = −27 ⇒ x y”, “x < y”, and “x = y” each fail because at least one admissible case contradicts them. “x ≥ y” also fails because x can be −27, which is less than y. Common Pitfalls: Forgetting that a squared equation has two roots; or treating √729 as ±27 (principal square root is +27 only). Mixing these causes incorrect certainty. Final Answer: Relationship cannot be determined

Q: Compare x and y from two quadratic statements (use mapping below): I. 2x^2 + 11x + 14 = 0 II. 4y^2 + 12y + 9 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context: We are asked to compare x and y when each variable is constrained by a quadratic equation. The key is to find all possible real values of x and y from their respective equations and check whether a single consistent inequality holds in all admissible cases. Given Data / Assumptions: I: 2x^2 + 11x + 14 = 0 II: 4y^2 + 12y + 9 = 0 Real roots are intended (these quadratics have real roots). Concept / Approach: Factor both quadratics to get explicit roots. Then compare every possible x from I with the unique y from II. If for all admissible x we have the same relation to y, the comparison is decided. Otherwise, it is indeterminate. Step-by-Step Solution: I: 2x^2 + 11x + 14 = (2x + 7)(x + 2) ⇒ x ∈ {−7/2, −2}II: 4y^2 + 12y + 9 = (2y + 3)^2 ⇒ y = −3/2Compare: −7/2 = −3.5 y” and “x = y” are contradicted by the computed values. “Relationship cannot be determined” is wrong because the relationship is the same in all admissible cases. Common Pitfalls: Missing the perfect square in II, or overlooking that I has two roots and both must be checked. Failing to test all cases can lead to incorrect indeterminacy. Final Answer: x < y

Q: Compare x and y given two quadratics (use mapping below): I. x^2 − 7x + 12 = 0 II. y^2 − 12y + 32 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context: Each variable is restricted by a quadratic with two possible roots. To decide a comparison between x and y, we must test all admissible combinations. If different admissible cases lead to different relations, then the overall comparison cannot be fixed. Given Data / Assumptions: I: x^2 − 7x + 12 = 0 II: y^2 − 12y + 32 = 0 Concept / Approach: Factor both quadratics. Enumerate all candidate values for x and y and compare. If at least one pair yields equality while another yields strict inequality, then a unique relation does not exist. Step-by-Step Solution: I factors: (x − 3)(x − 4) ⇒ x ∈ {3, 4}II factors: (y − 4)(y − 8) ⇒ y ∈ {4, 8}Test combinations:x = 3 with y = 4 or 8 ⇒ x y”, “x < y”, and “x = y” each fail for at least one admissible pair. Common Pitfalls: Choosing “x < y” based on one pair (e.g., 3 vs 8) and forgetting that equality also occurs (4 vs 4). All valid cases must be considered. Final Answer: Relationship cannot be determined

Q: Solve the simultaneous linear equations and compare x and y (use mapping below): I. 5x + 2y = 31 II. 3x + 7y = 36 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context: Two linear equations in two variables determine a unique solution pair (x, y). After solving, compare the numeric values directly to choose the correct relation using the provided mapping. Given Data / Assumptions: 5x + 2y = 31 3x + 7y = 36 Coefficients are such that a single unique solution exists. Concept / Approach: Use elimination. Align coefficients of y to subtract and solve for x, then back-substitute to find y. Finally, compare x and y numerically. Step-by-Step Solution: Multiply the first equation by 7: 35x + 14y = 217Multiply the second by 2: 6x + 14y = 72Subtract: (35x − 6x) + (14y − 14y) = 217 − 72 ⇒ 29x = 145 ⇒ x = 5Back-substitute: 5*5 + 2y = 31 ⇒ 25 + 2y = 31 ⇒ y = 3Comparison: x = 5 and y = 3 ⇒ x > y Verification / Alternative check: Plug x = 5, y = 3 into both equations to verify: 25 + 6 = 31 and 15 + 21 = 36. Both hold, confirming the unique pair. Why Other Options Are Wrong: “x y

Question 1

Form an equation with roots α/β and β/α: Given that α and β are roots of 2x^2 − 3x + 1 = 0, form the quadratic equation whose roots are α/β and β/α.

Accepted Answer

Introduction / Context: Transforming roots often relies on expressing new symmetric sums in terms of the old. If α and β satisfy a known quadratic, we can compute α + β and αβ, then find (α/β) + (β/α) and (α/β)(β/α) to construct the desired equation. Given Data / Assumptions: Original equation: 2x^2 − 3x + 1 = 0. Hence α + β = 3/2, αβ = 1/2. New roots: r1 = α/β and r2 = β/α. Concept / Approach: Compute sum and product of new roots: r1 + r2 = (α^2 + β^2)/(αβ) and r1*r2 = 1. Then use x^2 − (sum)x + (product) = 0 and clear denominators to get integer coefficients. Step-by-Step Solution: α^2 + β^2 = (α + β)^2 − 2αβ = (3/2)^2 − 2*(1/2) = 9/4 − 1 = 5/4.Sum r1 + r2 = (α^2 + β^2)/(αβ) = (5/4)/(1/2) = 5/2.Product r1*r2 = 1.Equation: x^2 − (5/2)x + 1 = 0 ⇒ multiply by 2 ⇒ 2x^2 − 5x + 2 = 0. Verification / Alternative check: Symmetry implies r1 and r2 are reciprocals; indeed product 1 confirms. Coefficients are integers after clearing denominators. Why Other Options Are Wrong: 2x^2 + 5x + 2 = 0 or 2x^2 − 5x − 2 = 0: Wrong sign patterns for sum and product. None of these: Incorrect since 2x^2 − 5x + 2 = 0 fits perfectly. Common Pitfalls: Miscomputing α^2 + β^2 or forgetting to divide by αβ. Ensure careful fraction arithmetic. Final Answer: 2x^2 − 5x + 2 = 0

Question 2

Factorization by pattern recognition: Factor the expression a^2 + 4b^2 + 4b − 4ab − 2a − 8 into a product of two linear factors.

Accepted Answer

Introduction / Context: We aim to factor a bivariate quadratic expression. Grouping terms or using a substitution like x = a − 2b to identify a perfect quadratic in x can simplify the process drastically, turning it into a one-variable factorization task. Given Data / Assumptions: Expression: a^2 + 4b^2 + 4b − 4ab − 2a − 8. Real a, b. Concept / Approach: Note that (a − 2b)^2 = a^2 − 4ab + 4b^2 appears within the expression. Set x = a − 2b and rewrite the expression in terms of x, then factor the resulting quadratic in x. Step-by-Step Solution: Rewrite: a^2 + 4b^2 − 4ab = (a − 2b)^2.Thus the expression becomes (a − 2b)^2 − 2a + 4b − 8.But a = (a − 2b) + 2b, so −2a + 4b = −2(a − 2b) − 0.Let x = a − 2b. Then expression = x^2 − 2x − 8 = (x − 4)(x + 2).Substitute back: (a − 2b − 4)(a − 2b + 2). Verification / Alternative check: Expand (a − 2b − 4)(a − 2b + 2) = (a − 2b)^2 − 2(a − 2b) − 8 = a^2 + 4b^2 − 4ab − 2a + 4b − 8, matching the original expression exactly. Why Other Options Are Wrong: Other pairs expand to different cross terms (e.g., +4ab instead of −4ab, or wrong linear coefficients), not reproducing the original expression. Common Pitfalls: Incorrect grouping leading to sign errors, or choosing x = a + 2b (which yields the wrong middle terms). The substitution x = a − 2b fits the pattern perfectly. Final Answer: (a − 2b − 4)(a − 2b + 2)

Question 3

Infinite nested radical value: Evaluate √(30 + √(30 + √(30 + …))) to its exact finite value.

Accepted Answer

Introduction / Context: Infinite nested radicals often converge to a finite value x that satisfies a self-referential equation: x = √(constant + x). Solving the resulting quadratic yields the limit value. Only the positive root is meaningful because the radical is nonnegative. Given Data / Assumptions: Expression: x = √(30 + x). x ≥ 0 due to square roots. Concept / Approach: Square both sides to eliminate the outer radical and form a quadratic in x. Solve and select the nonnegative solution consistent with the original expression. Step-by-Step Solution: Assume limit x exists with x = √(30 + x).Square: x^2 = 30 + x ⇒ x^2 − x − 30 = 0.Solve: Discriminant D = 1 + 120 = 121 ⇒ √D = 11.x = [1 ± 11]/2 ⇒ x = 6 or x = −5.Reject negative root. Hence x = 6. Verification / Alternative check: Plug back: √(30 + 6) = √36 = 6, confirming consistency. Why Other Options Are Wrong: 5, 7: Do not satisfy x = √(30 + x). 3√10 ≈ 9.486..., far from the correct fixed point. Common Pitfalls: Keeping both quadratic roots without checking the radical’s nonnegativity constraint, or mishandling the square step. Final Answer: 6

Question 4

Consecutive odd numbers (sum of squares = 394): The sum of the squares of two consecutive natural odd numbers is 394. Find the sum of the two numbers.

Accepted Answer

Introduction / Context: If two consecutive odd numbers are n and n + 2, then their squares sum to a quadratic in n. Solving gives the actual numbers. The final request is the sum of the numbers, which can then be obtained easily once n is known. Given Data / Assumptions: Numbers: n and n + 2, both odd, both natural. n^2 + (n + 2)^2 = 394. Concept / Approach: Expand and solve for n via the quadratic formula; choose the positive root. Then compute their sum S = n + (n + 2) = 2n + 2. Step-by-Step Solution: n^2 + (n + 2)^2 = 394 ⇒ 2n^2 + 4n + 4 = 394.2n^2 + 4n − 390 = 0 ⇒ divide by 2 ⇒ n^2 + 2n − 195 = 0.Discriminant D = 4 + 780 = 784 ⇒ √D = 28.n = [−2 ± 28]/2 ⇒ n = 13 or n = −15 (reject negative).Numbers: 13 and 15; Sum S = 28. Verification / Alternative check: 13^2 + 15^2 = 169 + 225 = 394. The sum 28 follows immediately. Why Other Options Are Wrong: 24, 32, 40: Do not correspond to the valid pair (13, 15). Common Pitfalls: Assuming difference 1 rather than 2 for odd consecutive numbers or arithmetic slips in expanding squares. Final Answer: 28

Question 5

Reciprocal roots condition (standard form): For the quadratic ax^2 + bx + c = 0 with nonzero roots, the roots are reciprocals of each other if and only if:

Accepted Answer

Introduction / Context: If r and s are the roots of ax^2 + bx + c = 0, then r*s = c/a by Vieta. The roots are reciprocals of each other precisely when r*s = 1 and both are nonzero, which directly translates to c/a = 1 ⇒ c = a. This is a textbook criterion for reciprocal roots. Given Data / Assumptions: Quadratic: ax^2 + bx + c = 0. Roots are real or complex but nonzero. Concept / Approach: Use the product of roots formula. The reciprocity condition requires the product to equal 1, hence c/a = 1. Step-by-Step Solution: Product of roots = c/a.Reciprocal condition: r*s = 1 ⇒ c/a = 1 ⇒ c = a. Verification / Alternative check: Consider transformed polynomial with roots 1/r and 1/s. Its monic form relates coefficients by swapping a and c, reinforcing the symmetry when c = a. Why Other Options Are Wrong: a = b or b = c: Do not constrain the product to be 1. None of these: Incorrect; c = a is the correct and sufficient condition. Common Pitfalls: Overlooking the nonzero requirement (a root 0 breaks reciprocity) or confusing with the equal-roots discriminant condition. Final Answer: c = a

Question 6

Compare x and y (define the answer codes): I. x^2 − 4 = 0 II. y^2 + 6y + 9 = 0 Roots are real (any root from each). Choose: A) x > y  B) x < y  C) x = y  D) Relationship cannot be determined

Accepted Answer

Introduction / Context: The first quadratic yields two symmetric roots ±2, and the second is a perfect square with a repeated root at −3. We must compare any root x from I with any root y from II and select the correct relation that always holds. Given Data / Assumptions: I: x^2 − 4 = 0 ⇒ x = 2 or x = −2. II: y^2 + 6y + 9 = 0 ⇒ (y + 3)^2 = 0 ⇒ y = −3. Concept / Approach: List all possible x values and the (single) y value. Compare systematically. If the smallest possible x is still greater than y, then x > y regardless of choice. Step-by-Step Solution: Possible x: {−2, 2}; y = −3.Compare: −2 > −3 and 2 > −3.Thus, for any choice, x > y. Verification / Alternative check: Draw a number line: y fixed at −3; both x values lie to the right (−2 and 2), confirming x > y. Why Other Options Are Wrong: x < y or x = y: Neither occurs for any valid selection. Relationship cannot be determined: Determined here; inequality holds for all choices. Common Pitfalls: Missing that y is a repeated single value (−3) and assuming variability comparable to x. The certainty arises from the fixed y. Final Answer: x > y

Question 7

Comparison of x and y from two statements (use the mapping below): I.  x^2 = 729 II. y = √729 (principal square root)  Use this mapping for the options: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined from the information g…

Accepted Answer

Introduction / Context: This comparison problem provides two separate statements about x and y. You must deduce the relationship between x and y without assuming anything beyond standard definitions. The twist is that I gives a squared equation with two solutions, while II fixes y via a principal square root, which is always the non-negative value. Given Data / Assumptions: I: x^2 = 729 II: y = √729, the principal square root Principal square root √n is defined as the unique non-negative number whose square is n. Concept / Approach: From I, solve for x by taking square roots, remembering both positive and negative roots. From II, determine y directly using the principal root. Then test all possible values of x against y to see if a single strict relation (>, <, =) always holds. If different cases give different relations, the comparison is indeterminate. Step-by-Step Solution: From I: x^2 = 729 ⇒ x = ±27From II: y = √729 = 27 (principal root)Case 1: x = 27 ⇒ x = yCase 2: x = −27 ⇒ x < y Verification / Alternative check: The set of solutions to x^2 = 729 is exactly {−27, 27}. Since both are compatible with the data, and these two cases produce different relations (equality in one, “less than” in the other), there is no unique comparison that always holds. Why Other Options Are Wrong: “x > y”, “x < y”, and “x = y” each fail because at least one admissible case contradicts them. “x ≥ y” also fails because x can be −27, which is less than y. Common Pitfalls: Forgetting that a squared equation has two roots; or treating √729 as ±27 (principal square root is +27 only). Mixing these causes incorrect certainty. Final Answer: Relationship cannot be determined

Question 8

Compare x and y from two quadratic statements (use mapping below): I.  2x^2 + 11x + 14 = 0 II. 4y^2 + 12y + 9 = 0  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context: We are asked to compare x and y when each variable is constrained by a quadratic equation. The key is to find all possible real values of x and y from their respective equations and check whether a single consistent inequality holds in all admissible cases. Given Data / Assumptions: I: 2x^2 + 11x + 14 = 0 II: 4y^2 + 12y + 9 = 0 Real roots are intended (these quadratics have real roots). Concept / Approach: Factor both quadratics to get explicit roots. Then compare every possible x from I with the unique y from II. If for all admissible x we have the same relation to y, the comparison is decided. Otherwise, it is indeterminate. Step-by-Step Solution: I: 2x^2 + 11x + 14 = (2x + 7)(x + 2) ⇒ x ∈ {−7/2, −2}II: 4y^2 + 12y + 9 = (2y + 3)^2 ⇒ y = −3/2Compare: −7/2 = −3.5 < −1.5, and −2 < −1.5Thus, in all cases, x < y. Verification / Alternative check: Compute numeric values precisely and reconfirm: x candidates are −3.5 and −2; y is −1.5. Both comparisons show x is strictly less than y. Why Other Options Are Wrong: “x > y” and “x = y” are contradicted by the computed values. “Relationship cannot be determined” is wrong because the relationship is the same in all admissible cases. Common Pitfalls: Missing the perfect square in II, or overlooking that I has two roots and both must be checked. Failing to test all cases can lead to incorrect indeterminacy. Final Answer: x < y

Question 9

Compare x and y given two quadratics (use mapping below): I.  x^2 − 7x + 12 = 0 II. y^2 − 12y + 32 = 0  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context: Each variable is restricted by a quadratic with two possible roots. To decide a comparison between x and y, we must test all admissible combinations. If different admissible cases lead to different relations, then the overall comparison cannot be fixed. Given Data / Assumptions: I: x^2 − 7x + 12 = 0 II: y^2 − 12y + 32 = 0 Concept / Approach: Factor both quadratics. Enumerate all candidate values for x and y and compare. If at least one pair yields equality while another yields strict inequality, then a unique relation does not exist. Step-by-Step Solution: I factors: (x − 3)(x − 4) ⇒ x ∈ {3, 4}II factors: (y − 4)(y − 8) ⇒ y ∈ {4, 8}Test combinations:x = 3 with y = 4 or 8 ⇒ x < yx = 4 with y = 4 ⇒ x = yx = 4 with y = 8 ⇒ x < y Verification / Alternative check: Because both x = 4, y = 4 (equality) and x = 3, y = 4 (strict inequality x < y) are valid, there is no single definitive comparison valid for all admissible cases. Why Other Options Are Wrong: “x > y”, “x < y”, and “x = y” each fail for at least one admissible pair. Common Pitfalls: Choosing “x < y” based on one pair (e.g., 3 vs 8) and forgetting that equality also occurs (4 vs 4). All valid cases must be considered. Final Answer: Relationship cannot be determined

Question 10

Solve the simultaneous linear equations and compare x and y (use mapping below): I.  5x + 2y = 31 II. 3x + 7y = 36  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context: Two linear equations in two variables determine a unique solution pair (x, y). After solving, compare the numeric values directly to choose the correct relation using the provided mapping. Given Data / Assumptions: 5x + 2y = 31 3x + 7y = 36 Coefficients are such that a single unique solution exists. Concept / Approach: Use elimination. Align coefficients of y to subtract and solve for x, then back-substitute to find y. Finally, compare x and y numerically. Step-by-Step Solution: Multiply the first equation by 7: 35x + 14y = 217Multiply the second by 2: 6x + 14y = 72Subtract: (35x − 6x) + (14y − 14y) = 217 − 72 ⇒ 29x = 145 ⇒ x = 5Back-substitute: 5*5 + 2y = 31 ⇒ 25 + 2y = 31 ⇒ y = 3Comparison: x = 5 and y = 3 ⇒ x > y Verification / Alternative check: Plug x = 5, y = 3 into both equations to verify: 25 + 6 = 31 and 15 + 21 = 36. Both hold, confirming the unique pair. Why Other Options Are Wrong: “x < y” and “x = y” contradict the computed values; “Relationship cannot be determined” is invalid because the system gives a unique, checkable solution. Common Pitfalls: Arithmetic slips when multiplying or subtracting the equations; recheck the elimination step carefully. Final Answer: x > y

Question 11

Condition for exactly one real (repeated) root of x^2 − p x + q = 0 (with p, q ∈ ℝ): State the correct discriminant condition ensuring a single real root.

Accepted Answer

Introduction / Context: A quadratic equation ax^2 + bx + c = 0 has its number and nature of real roots governed by the discriminant D = b^2 − 4ac. Here a = 1, b = −p, c = q. The phrase “exactly one real root” means a repeated real root (a double root), which occurs precisely when the discriminant equals zero. Given Data / Assumptions: Equation: x^2 − p x + q = 0 Parameters: p, q ∈ ℝ, with leading coefficient 1 ≠ 0 We seek the condition for one real (repeated) root. Concept / Approach: Use the discriminant test. For a quadratic ax^2 + bx + c = 0: D > 0 ⇒ two distinct real roots; D = 0 ⇒ one real repeated root; D < 0 ⇒ no real roots (complex conjugate pair). Identify b and c in terms of p and q, compute D, and enforce D = 0. Step-by-Step Solution: Here a = 1, b = −p, c = qDiscriminant D = b^2 − 4ac = (−p)^2 − 4*1*q = p^2 − 4qExactly one real root ⇔ D = 0 ⇔ p^2 − 4q = 0Therefore the required condition is p^2 = 4q Verification / Alternative check: When p^2 = 4q, the quadratic becomes (x − p/2)^2 = 0 after completing the square, which clearly has a single repeated root at x = p/2. Why Other Options Are Wrong: p^2 < 4q gives D < 0 (no real roots). p^2 > 4q gives D > 0 (two distinct real roots). “No condition” and “p^2 ≤ 4q” do not specifically guarantee exactly one real root. Common Pitfalls: Confusing “at least one real root” (D ≥ 0) with “exactly one real root” (D = 0). The equality case is essential here. Final Answer: p^2 = 4q

Question 12

Parameter k when x = 3 is a root of 3x^2 + (k − 1)x + 9 = 0. Find the value of k that satisfies this condition.

Accepted Answer

Introduction / Context: When a number r is a root of a polynomial, substituting x = r makes the polynomial equal zero. Here we are told x = 3 is a root of a quadratic with unknown parameter k, and we must determine k so that the equation holds true. Given Data / Assumptions: Quadratic: 3x^2 + (k − 1)x + 9 = 0 Condition: x = 3 is a solution Concept / Approach: Use the root-substitution property: plug x = 3 into the left-hand side, set the result to zero, and solve the resulting linear equation in k. Step-by-Step Solution: Substitute x = 3: 3*(3^2) + (k − 1)*3 + 9 = 0Compute: 3*9 + 3k − 3 + 9 = 0 ⇒ 27 + 3k − 3 + 9 = 0Combine constants: 27 − 3 + 9 = 33 ⇒ 3k + 33 = 0Solve: 3k = −33 ⇒ k = −11 Verification / Alternative check: Plug k = −11 back: 3x^2 + (−12)x + 9 = 0. For x = 3: 27 − 36 + 9 = 0, verified. Why Other Options Are Wrong: 13, −13, 11 do not satisfy 3k + 33 = 0; only k = −11 works. “−9” is a distractor without algebraic support. Common Pitfalls: Arithmetic slips when adding constants after substitution. Carefully add 27, −3, and 9 to avoid errors. Final Answer: −11

Question 13

Given one root of 3x^2 − 10x + 3 = 0 is 1/3. Find the other root of the quadratic.

Accepted Answer

Introduction / Context: For a quadratic ax^2 + bx + c = 0, the sum and product of roots are given by S = −b/a and P = c/a. Knowing one root allows quick computation of the other using the product (or sum). Given Data / Assumptions: Quadratic: 3x^2 − 10x + 3 = 0 One root r1 = 1/3 Concept / Approach: Use product of roots P = c/a. If r1 is known, then r2 = P / r1. This avoids solving the quadratic again. Optionally, confirm by substitution or by using the sum of roots formula. Step-by-Step Solution: a = 3, b = −10, c = 3 ⇒ P = c/a = 3/3 = 1Given r1 = 1/3, then r2 = P / r1 = 1 / (1/3) = 3 Verification / Alternative check: Check via substitution: For x = 1/3, LHS = 3*(1/9) − 10*(1/3) + 3 = 1/3 − 10/3 + 3 = 0. For x = 3, LHS = 27 − 30 + 3 = 0. Both satisfy the equation. Why Other Options Are Wrong: 1/3 is the given root, not the other; −3 and 10/3 do not satisfy the equation as roots; “None of these” is unnecessary as a correct value exists among options. Common Pitfalls: Using the sum instead of product incorrectly or mishandling fractions. Product method is clean and fast here. Final Answer: 3

Question 14

Factor the biquadratic expression x^4 + 7x^2 + 16 over the reals.

Accepted Answer

Introduction / Context: Biquadratic expressions in x^4 can often be factored by treating x^2 as a single variable t. The goal is to express x^4 + 7x^2 + 16 as a product of two real quadratics with zero x^3 and x terms so that expansion matches the original middle term 7x^2 and constant 16. Given Data / Assumptions: Expression: x^4 + 7x^2 + 16 We seek a factorization over the reals. Concept / Approach: Assume a symmetric factorization: (x^2 + ax + 4)(x^2 + bx + 4). To eliminate x^3 and x terms, choose b = −a. Then the x^2 coefficient becomes (ab + 8) = (−a^2 + 8). Match this to 7 to solve for a. Step-by-Step Solution: Let factors be (x^2 + ax + 4)(x^2 − ax + 4)Expand: x^4 + (ab + 8)x^2 + 16 with ab = −a^2Coefficient of x^2: −a^2 + 8 must equal 7 ⇒ a^2 = 1 ⇒ a = ±1Thus a = 1 gives (x^2 + x + 4)(x^2 − x + 4) Verification / Alternative check: Expand (x^2 + x + 4)(x^2 − x + 4): you get x^4 + 7x^2 + 16 exactly, with x^3 and x coefficients canceling due to symmetry. Why Other Options Are Wrong: Other factorizations either create nonzero x^3/x terms or yield incorrect x^2 coefficients/constant terms; “Irreducible over reals” is incorrect since a correct real factorization exists. Common Pitfalls: Factoring as a quadratic in x^2 only (which has negative discriminant) and concluding irreducible; but symmetric real quadratic factors still exist. Final Answer: (x^2 + x + 4)(x^2 - x + 4)

Question 15

Find the common root of the quadratics x^2 − 7x + 10 = 0 and x^2 − 10x + 16 = 0.

Accepted Answer

Introduction / Context: Two quadratics are given. A common root is a value of x that satisfies both equations. Factoring each quadratic allows direct identification of roots and their intersection. Given Data / Assumptions: x^2 − 7x + 10 = 0 x^2 − 10x + 16 = 0 Concept / Approach: Factor both polynomials into linear factors and list their roots. The common root is the value appearing in both root sets. Step-by-Step Solution: x^2 − 7x + 10 = (x − 5)(x − 2) ⇒ roots: {5, 2}x^2 − 10x + 16 = (x − 8)(x − 2) ⇒ roots: {8, 2}Common root = 2 Verification / Alternative check: Plug x = 2 into both equations: 4 − 14 + 10 = 0 and 4 − 20 + 16 = 0. Both equal zero, confirming. Why Other Options Are Wrong: −2, 3, 5, 8 are not roots common to both equations; only 2 appears in both sets. Common Pitfalls: Arithmetic slips in factoring; check factor pairs that sum to the middle coefficient and multiply to the constant term. Final Answer: 2

Question 16

Divide 16 into two parts so that twice the square of the larger exceeds the square of the smaller by 164. Find the two parts (larger, smaller).

Accepted Answer

Introduction / Context: This is a quadratic modeling question. Let the two parts be a (larger) and b (smaller) with a + b = 16. The condition links their squares: 2a^2 − b^2 = 164. Solve the system to identify the two numbers. Given Data / Assumptions: a + b = 16 (a ≥ b ≥ 0 implicit) 2a^2 − b^2 = 164 Concept / Approach: Express b as 16 − a and substitute into the square relation to obtain a single quadratic in a. Solve it, pick the physically meaningful root (nonnegative and larger), and then find b from the sum constraint. Step-by-Step Solution: Let b = 16 − aCondition: 2a^2 − (16 − a)^2 = 164Expand: 2a^2 − (256 − 32a + a^2) = 164 ⇒ a^2 + 32a − 256 = 164Simplify: a^2 + 32a − 420 = 0Discriminant: 32^2 + 4*420 = 1024 + 1680 = 2704 ⇒ √2704 = 52a = (−32 ± 52)/2 ⇒ a = 10 or a = −42 (reject)Thus a = 10; b = 16 − 10 = 6 Verification / Alternative check: Check: 2*(10^2) − 6^2 = 200 − 36 = 164, satisfied. Sum 10 + 6 = 16, satisfied. Why Other Options Are Wrong: 8,8 gives 2*64 − 64 = 64, not 164; 12,4 gives 288 − 16 = 272; 11,5 gives 242 − 25 = 217. Hence only 10,6 works. Common Pitfalls: Sign errors when expanding (16 − a)^2 or dropping the negative root without checking feasibility. Always verify both constraints. Final Answer: 10, 6

Question 17

Solve the logarithmic quadratic: If log10(x^2 − 6x + 45) = 2, find all real values of x.

Accepted Answer

Introduction / Context: We are given a base-10 logarithm of a quadratic expression equal to 2. Use the definition of logarithms to convert to an exponential equation and solve the resulting quadratic. Finally, confirm that the argument of the logarithm is positive for each solution (domain check). Given Data / Assumptions: log10(x^2 − 6x + 45) = 2 Domain requirement: x^2 − 6x + 45 > 0 Concept / Approach: Use a^b = 10^2 when log10(a) = 2. So set the quadratic equal to 100 and solve. Because the resulting quadratic will have real roots, check that each keeps the log argument positive (which it does here since the expression equals 100 at the roots and is a parabola opening upward). Step-by-Step Solution: x^2 − 6x + 45 = 10^2 = 100x^2 − 6x − 55 = 0Discriminant: 36 + 220 = 256 ⇒ √256 = 16x = (6 ± 16)/2 ⇒ x = 11 or x = −5 Verification / Alternative check: At x = 11: x^2 − 6x + 45 = 121 − 66 + 45 = 100 ⇒ log10(100) = 2. At x = −5: 25 + 30 + 45 = 100 ⇒ log10(100) = 2. Domain satisfied (argument = 100 > 0). Why Other Options Are Wrong: 6,9 and 10,5 do not satisfy the transformed quadratic; “Only 11” ignores the second valid root; 9, −5 has 9 incorrect. Common Pitfalls: Failing to enforce the log domain or miscomputing the discriminant. Always convert log equations carefully and check positivity. Final Answer: 11, −5

Question 18

Two quadratics share a common root: x^2 + 2x − 3 = 0 and x^2 + 3x − k = 0. Find the non-zero value of k.

Accepted Answer

Introduction / Context: If two quadratics share a root r, then r satisfies both equations. Solve the first to get its roots, test them in the second, and find k that makes the second equation zero for that r. The question specifies the non-zero k if there is a choice. Given Data / Assumptions: Eqn1: x^2 + 2x − 3 = 0 Eqn2: x^2 + 3x − k = 0 Common real root exists; choose non-zero k. Concept / Approach: Factor Eqn1 to get candidate roots. Substitute each into Eqn2 to solve for k. If multiple k values occur, follow the “non-zero” instruction. Step-by-Step Solution: Eqn1 factors: (x + 3)(x − 1) = 0 ⇒ roots r ∈ {−3, 1}Substitute r = 1 into Eqn2: 1 + 3 − k = 0 ⇒ k = 4Substitute r = −3 into Eqn2: 9 − 9 − k = 0 ⇒ k = 0 (disallowed as per prompt)Hence the non-zero k is 4 Verification / Alternative check: With k = 4, Eqn2 becomes x^2 + 3x − 4 = 0 = (x + 4)(x − 1); root x = 1 matches Eqn1. Why Other Options Are Wrong: 1, 2, 3 do not produce a shared root with Eqn1; only 4 works and is non-zero. Common Pitfalls: Missing the “non-zero” qualifier and selecting k = 0. Always read such constraints carefully. Final Answer: 4

Question 19

Find the quadratic whose roots are reciprocals of the roots of 3x^2 − 20x + 17 = 0.

Accepted Answer

Introduction / Context: If α and β are roots of a quadratic, then 1/α and 1/β are roots of another quadratic obtained by the substitution x → 1/x and clearing denominators. This is a standard transformation for reciprocal roots. Given Data / Assumptions: Original quadratic: 3x^2 − 20x + 17 = 0 We need the equation with roots 1/α and 1/β Concept / Approach: Replace x by 1/x: 3*(1/x^2) − 20*(1/x) + 17 = 0. Multiply both sides by x^2 to clear denominators and obtain a monic-in-form quadratic in x (up to a constant factor). Step-by-Step Solution: Start: 3/x^2 − 20/x + 17 = 0Multiply by x^2: 3 − 20x + 17x^2 = 0Reorder: 17x^2 − 20x + 3 = 0 Verification / Alternative check: By Vieta: For original, sum = 20/3 and product = 17/3. For reciprocal equation, sum = (α+β)/(αβ) = (20/3)/(17/3) = 20/17, product = 1/(αβ) = 3/17, matching coefficients of 17x^2 − 20x + 3 = 0 after normalization. Why Other Options Are Wrong: Signs or constants are incorrect for a reciprocal transformation; only option (a) matches the derived equation. Common Pitfalls: Forgetting to multiply through by x^2 or mishandling the order/signs when rewriting. Final Answer: 17x^2 − 20x + 3 = 0

Question 20

Nature of factors for x^2 − x + 1 over the reals: determine whether proper linear factors exist.

Accepted Answer

Introduction / Context: The factorability of a quadratic over the reals depends on its discriminant. If the discriminant is negative, the quadratic has no real roots and therefore cannot be factored into real linear factors. We analyze x^2 − x + 1 accordingly. Given Data / Assumptions: Quadratic: x^2 − x + 1 Coefficients are real. Concept / Approach: Compute the discriminant D = b^2 − 4ac for a = 1, b = −1, c = 1. Interpret D: D > 0 ⇒ two distinct real linear factors; D = 0 ⇒ a repeated real linear factor; D < 0 ⇒ no real linear factors. Step-by-Step Solution: a = 1, b = −1, c = 1D = (−1)^2 − 4*1*1 = 1 − 4 = −3Since D < 0, there are no real roots and no factorization into real linear factors. Verification / Alternative check: Graph y = x^2 − x + 1 has vertex above the x-axis and never crosses, confirming no real zeros. Why Other Options Are Wrong: “One/two proper linear factor(s)” require D ≥ 0; “exactly one repeated linear factor” corresponds to D = 0, which is not the case; “None of these” is unnecessary. Common Pitfalls: Miscomputing D or assuming any quadratic factors over the reals. Always check the discriminant first. Final Answer: no proper linear factor

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