Q: Let x be the greater real root of x^2 − 365 = 364 and y satisfy y − √324 = √81. Compare x and y.

Introduction / Context: We compare exact numerical values derived from simple square roots. The equations are arranged to yield integers after evaluation, simplifying the comparison. Given Data / Assumptions: x^2 − 365 = 364 ⇒ x^2 = 729 ⇒ roots ±27; greater x = 27. y − √324 = √81 ⇒ y − 18 = 9 ⇒ y = 27. Concept / Approach: Evaluate each side carefully. √324 = 18 and √81 = 9. Then solve the simple linear relation for y and compare with x. Step-by-Step Solution: x = 27.y = 27.Therefore x = y, which implies x ≥ y (and also x ≤ y). Verification / Alternative check: Substitutions confirm both equations are satisfied by 27. Why Other Options Are Wrong: Strict inequalities do not hold since equality is exact. Common Pitfalls: Misreading the radicals or arithmetic around subtracting and adding 18 and 9. Final Answer: If x ≥ y

Question 1

Let x be the greater real root of x^2 − x − 12 = 0 and y be the greater real root of y^2 + 5y + 6 = 0. Compare x and y.

Accepted Answer

Introduction / Context: Straightforward factoring reveals both sets of roots, and the greater ones are compared directly as per the clarified convention. Given Data / Assumptions: x^2 − x − 12 = 0 ⇒ (x − 4)(x + 3) = 0 ⇒ greater x = 4. y^2 + 5y + 6 = 0 ⇒ (y + 2)(y + 3) = 0 ⇒ greater y = −2. Concept / Approach: Factor both quadratics. Identify greater real roots. Compare numerically. Step-by-Step Solution: x = 4; y = −2.Hence x > y. Verification / Alternative check: Plugging back confirms each value solves the respective equation. Why Other Options Are Wrong: They contradict 4 > −2. Common Pitfalls: Picking the smaller root for either quadratic or sign mistakes during factoring. Final Answer: If x > y

Question 2

Let x be the greater real root of 3x^2 + 8x + 4 = 0 and y be the greater real root of 4y^2 − 19y + 12 = 0. Compare x and y.

Accepted Answer

Introduction / Context: Here we compare greater roots from two non-monic quadratics. Using the quadratic formula (or completing the square) gives exact values for comparison. Given Data / Assumptions: 3x^2 + 8x + 4 = 0. 4y^2 − 19y + 12 = 0. Concept / Approach: Compute each pair of roots. Identify the larger root in each case and compare numerically. Step-by-Step Solution: For x: Δ = 8^2 − 4*3*4 = 64 − 48 = 16; roots = (−8 ± 4)/(2*3) = {−2, −2/3}. Greater x = −2/3.For y: Δ = (−19)^2 − 4*4*12 = 361 − 192 = 169; roots = (19 ± 13)/(8) = {4, 3/4}. Greater y = 4.Thus x = −2/3 and y = 4 ⇒ x < y. Verification / Alternative check: Quick decimal comparison (−0.666… vs 4) confirms the inequality. Why Other Options Are Wrong: They contradict the numerical ordering. Common Pitfalls: Mixing up which of the two y-roots is larger or arithmetic slips in discriminant calculations. Final Answer: If x < y

Question 3

Let x be the greater real root of x^2 − 365 = 364 and y satisfy y − √324 = √81. Compare x and y.

Accepted Answer

Introduction / Context: We compare exact numerical values derived from simple square roots. The equations are arranged to yield integers after evaluation, simplifying the comparison. Given Data / Assumptions: x^2 − 365 = 364 ⇒ x^2 = 729 ⇒ roots ±27; greater x = 27. y − √324 = √81 ⇒ y − 18 = 9 ⇒ y = 27. Concept / Approach: Evaluate each side carefully. √324 = 18 and √81 = 9. Then solve the simple linear relation for y and compare with x. Step-by-Step Solution: x = 27.y = 27.Therefore x = y, which implies x ≥ y (and also x ≤ y). Verification / Alternative check: Substitutions confirm both equations are satisfied by 27. Why Other Options Are Wrong: Strict inequalities do not hold since equality is exact. Common Pitfalls: Misreading the radicals or arithmetic around subtracting and adding 18 and 9. Final Answer: If x ≥ y

Question 4

Solve and compare: Let x satisfy (4/√x) + (7/√x) = √x with x > 0, and let y satisfy y^2 − (11)^(5/2)/√y = 0 with y > 0. Compare x and y.

Accepted Answer

Introduction / Context: Two radical equations are posed with positivity constraints (because of square roots). We solve each exactly by appropriate substitutions and then compare the resulting values. Given Data / Assumptions: (4/√x) + (7/√x) = √x with x > 0. y^2 − (11)^(5/2)/√y = 0 with y > 0. Concept / Approach: For x, combine like terms: (11/√x) = √x ⇒ cross-multiply to get x. For y, substitute t = √y (t > 0) to reduce to a simple power equation. Step-by-Step Solution: (11/√x) = √x ⇒ multiply both sides by √x: 11 = x ⇒ x = 11.For y: let t = √y ⇒ y = t^2. Equation: t^4 − (11)^(5/2)/t = 0.Multiply by t: t^5 = (11)^(5/2).Take positive fifth root: t = (11)^(1/2) ⇒ √y = √11 ⇒ y = 11.Thus x = y = 11, which satisfies x ≥ y. Verification / Alternative check: Substitute back to confirm each equation balances at 11. Why Other Options Are Wrong: Strict inequalities are incorrect; equality holds exactly. Common Pitfalls: Mishandling fractional exponents, or allowing negative square-root values contrary to domain constraints. Final Answer: if x ≥ y

Question 5

Tour-budget planning: Mr. Arjun has ₹ 360 for his trip. If he extends his tour by 4 days, he must reduce his daily expense by ₹ 3 to stay within budget. For how many days was the original plan?

Accepted Answer

Introduction / Context: This is a classic unitary-method / algebra blend about fixed total budget and varying duration. Extending the number of days forces a reduction in daily spending so that the total cost remains the same. We form an equation in the number of days and solve it cleanly. Given Data / Assumptions: Total budget B = ₹ 360. Original number of days = d; original daily expense e = 360/d. Extended trip: d + 4 days; new daily expense e − 3 = 360/(d + 4). d is a positive integer. Concept / Approach: Equate the two expressions for the new daily expense. This yields a rational equation in d that simplifies to a quadratic. Solve for d and choose the positive, meaningful value. Step-by-Step Solution: Original daily expense e = 360/d.New daily expense e − 3 = 360/(d + 4).So 360/d − 3 = 360/(d + 4).Bring to one side: 360/d − 360/(d + 4) = 3.Compute LHS: 360[(d + 4) − d] / (d(d + 4)) = 1440 / (d(d + 4)).Thus 1440 / (d(d + 4)) = 3 ⇒ d(d + 4) = 480.d^2 + 4d − 480 = 0 ⇒ Discriminant = 16 + 1920 = 1936 = 44^2.d = (−4 + 44)/2 = 20 (take positive root). Verification / Alternative check: Original e = 360/20 = ₹ 18/day. New duration 24 days with ₹ 15/day gives total 24*15 = ₹ 360, consistent. Why Other Options Are Wrong: 40, 60, 15, 24 do not satisfy the budget relation; only d = 20 preserves the total expense under the given change. Common Pitfalls: Algebra slips when clearing denominators; forgetting that total budget is fixed in both scenarios. Final Answer: 20

Question 6

Quadratic with rational coefficients: If a quadratic equation with rational coefficients has one root equal to √5, then which equation below can it be?

Accepted Answer

Introduction / Context: This question checks your understanding of conjugate roots for polynomials with rational (or real) coefficients. If an irrational root like √5 occurs, its conjugate −√5 must also be a root so that the product expands to a polynomial with rational coefficients. Given Data / Assumptions: One root is √5. Coefficients are rational numbers. We need a quadratic (degree 2) that admits √5 as a root and has only rational coefficients. Concept / Approach: For quadratics with rational coefficients, irrational roots appear in conjugate pairs. So if √5 is a root, then −√5 must also be a root. The minimal monic polynomial having roots ±√5 is (x − √5)(x + √5) = x^2 − 5. Step-by-Step Solution: Assume roots are r1 = √5 and r2 = −√5. Form the quadratic: (x − r1)(x − r2) = (x − √5)(x + √5) = x^2 − (√5)^2 = x^2 − 5. Hence, a valid quadratic with rational coefficients is x^2 − 5 = 0. Verification / Alternative check: Plug x = √5 into x^2 − 5: (√5)^2 − 5 = 5 − 5 = 0. Works. Plug x = −√5: (−√5)^2 − 5 = 5 − 5 = 0. Also works. Why Other Options Are Wrong: x^2 + 5 = 0 has imaginary roots ±i√5; x^2 − 10 = 0 has roots ±√10 (not √5); the extra option with √5 in the coefficient is not rational-coefficient; “None of these” is incorrect because x^2 − 5 = 0 fits perfectly. Common Pitfalls: Assuming √5 alone can be a root without its conjugate for rational coefficients, or confusing √5 with √10 or i√5. Final Answer: x^2 − 5 = 0

Question 7

Recover and solve the intended equation (repair applied): Interpret the corrupted stem as ( (x + 4) / (x − 4) ) + ( (x − 4) / (x + 4) ) = 10 / 3. Find all real roots x.

Accepted Answer

Introduction / Context: The original database text was garbled. Using the Recovery-First Policy, we interpret it as a classic rational-expression identity: (x + 4)/(x − 4) + (x − 4)/(x + 4) = 10/3. This kind of expression simplifies via a common denominator, producing an equation only in x^2. We then solve for x and check domain restrictions (x ≠ ±4). Given Data / Assumptions: (x + 4)/(x − 4) + (x − 4)/(x + 4) = 10/3. Domain: x ≠ 4, x ≠ −4 (to avoid zero denominators). Concept / Approach: Use the identity (A/B) + (B/A) = (A^2 + B^2)/(AB). Here, A = x + 4 and B = x − 4. This yields a rational equation in x that simplifies to a quadratic in x^2. Solve and select the real roots allowed by the domain. Step-by-Step Solution: (x + 4)/(x − 4) + (x − 4)/(x + 4) = [(x + 4)^2 + (x − 4)^2]/(x^2 − 16) Numerator = (x^2 + 8x + 16) + (x^2 − 8x + 16) = 2x^2 + 32 So (2x^2 + 32)/(x^2 − 16) = 10/3 Cross-multiply: 3(2x^2 + 32) = 10(x^2 − 16) ⇒ 6x^2 + 96 = 10x^2 − 160 Rearrange: 4x^2 = 256 ⇒ x^2 = 64 ⇒ x = ±8 Verification / Alternative check: Denominator check: x ≠ ±4, so ±8 are valid. Substitute x = 8 (or −8) to confirm the left-hand side equals 10/3 numerically. Why Other Options Are Wrong: ±4 are excluded (division by zero). ±6 and 2 ± √3 do not satisfy the equation when substituted. “No real root” is false since ±8 work. Common Pitfalls: Dropping the domain restriction, or algebra errors when simplifying the rational sum can lead to spurious answers. Final Answer: ± 8

Question 8

Common roots across two quadratics: I. x^2 − 19x + 84 = 0 II. y^2 − 25y + 156 = 0 How many integer values are common to the two root-sets?

Accepted Answer

Introduction / Context: Instead of comparing unknowns, we repair the ambiguous stem to a precise and valuable task: count the common integer roots of two given quadratics. Factoring both equations quickly reveals the root sets and their intersection size. Given Data / Assumptions: I: x^2 − 19x + 84 = 0 II: y^2 − 25y + 156 = 0 We count how many integers appear in both sets of roots. Concept / Approach: Factor each quadratic: for t^2 − St + P = 0, find two integers whose sum is S and product is P. The solutions are those integers. Then take the intersection of the two sets and count elements. Step-by-Step Solution: I: x^2 − 19x + 84 = 0 ⇒ (x − 12)(x − 7) = 0 ⇒ roots {12, 7} II: y^2 − 25y + 156 = 0 ⇒ (y − 12)(y − 13) = 0 ⇒ roots {12, 13} Intersection = {12}; count = 1 Verification / Alternative check: Check by substitution: each candidate in the intersection should satisfy both quadratics. 12 satisfies both; 7 and 13 are unique to one each. Why Other Options Are Wrong: 2, 3, 4, 0 miscount the intersection size; only one integer (12) is shared. Common Pitfalls: Arithmetic slips while factoring, or missing that “how many” asks for count of shared values, not listing all roots. Final Answer: 1

Question 9

Ⅰ. 12 - 23 = 5√x √ x √ x Ⅱ. √ y - 5 √ y = 1 12 12 √ y

Accepted Answer

Question 10

Ⅰ. 9 + 19 = √x √ x √ x Ⅱ. y5 - ( 2 × 14 ) 11/2 = 0 √ y

Accepted Answer

Question 11

Ⅰ. √ 784x + 1234 = 1486 Ⅱ. √ 1089y + 2081 = 2345

Accepted Answer

According to question ,we can say that From equation Ⅰ. √784x + 1234 = 1486 ⇒ √784x = 252 ⇒ 28x = 252 ⇒ x = 9 From equation Ⅱ. √1089y + 2081 = 2345 ⇒ √1089y = 264 ⇒ 33y = 264 ⇒ y = 8 ∴ x > y Thus , required answer will be x > y .

Question 12

Ⅰ. x 3 – 468 = 1729 Ⅱ. y 2 – 1733 + 1564 = 0

Accepted Answer

According to question, From equation Ⅰ. x3 – 468 = 1729 ⇒ x3 = 2197 = 133 ⇒ x = 13 From equation Ⅱ. y2 – 1733 + 1564 = 0 ⇒ y2 = 169 = 132 ⇒ y = ±13 Thus , x ≥ y is required answer .

Question 13

Ⅰ. x 2 – 7x + 12 = 0 Ⅱ. y 2 + y – 12 = 0

Accepted Answer

From the given equations , we can say that From equation Ⅰ. x2 - 7x + 12 = 0 ⇒ x2 - 4x - 3x + 12 = 0 ⇒ x( x - 4 ) - 3( x - 4 ) = 0 ⇒ ( x - 3 )( x - 4 ) = 0 ⇒ x = 3 or x = 4 From equation Ⅱ. y2 + y - 12 = 0 ⇒ y2 + 4y - 3y - 12 = 0 ⇒ y( y + 4 ) - 3( y + 4 ) = 0 ⇒ ( y - 3 )( y + 4 ) = 0 ⇒ y = 3 or y = - 4 From above both equations it is clear that x ≥ y is correct answer .

Question 14

Ⅰ. x 4 – 227 = 398 Ⅱ. y 2 + 321 = 346

Accepted Answer

According to question ,we can say that From equation Ⅰ. x4 = 398 + 227 = 625 ⇒ x4 = 54 From equation Ⅱ. y2 = 346 - 321 = 25 ⇒ y2 = 52

Question 15

Ⅰ. x 2 - 1 = 0 Ⅱ. y 2 + 4y + 3 = 0

Accepted Answer

From the given equations , we know that Ⅰ. x2 - 1 = 0 ⇒ x2 = 1 ⇒ x = ± √1 = ± 1 Ⅱ. y2 + 4y + 3 = 0 ⇒ y2 + y + 3y + 3 = 0 ⇒ y( y + 1) + 3( y + 1) = 0 ⇒ ( y + 1) ( y + 3) ⇒ y = -3 or -1 From above equations it is clear that x ≥ y is correct answer .

Question 16

Ⅰ. x 3 - 371 = 629 Ⅱ. y 3 - 543 = 788

Accepted Answer

According to given question , we have Ⅰ. x3 - 371 = 629 ⇒ x = √1000 = 10 Ⅱ. y3 - 543 = 788 ⇒ y = √1331 = 11 From above both equations it is clear that x < y is correct answer .

Question 17

Ⅰ. 2x 2 + 11x + 12 = 0 Ⅱ. 5y 2 + 27y + 10 = 0

Accepted Answer

Question 18

Known root in a quadratic, find the other root and coefficient: Given 2x^2 + P x + 4 = 0 has a root x = 2. Find the second root and the value of P.

Accepted Answer

Introduction / Context: When one root of a quadratic is known, you can plug it into the equation to determine the missing coefficient. Then use Vieta’s relations (sum and product of roots) to find the second root efficiently without re-solving the entire quadratic from scratch. Given Data / Assumptions: Quadratic: 2x^2 + P x + 4 = 0 One root r1 = 2. Concept / Approach: Substitute r1 into the polynomial to determine P. Then apply sum of roots = −P/2 and product of roots = 4/2 = 2 to deduce the second root r2. Cross-check both sum and product to avoid slips. Step-by-Step Solution: Plug x = 2: 2*(2^2) + P*2 + 4 = 0 ⇒ 8 + 2P + 4 = 0 ⇒ 2P = −12 ⇒ P = −6 Sum of roots = −P/2 = −(−6)/2 = 3 Let the second root be r2. Then 2 + r2 = 3 ⇒ r2 = 1 Verification / Alternative check: Product = c/a = 4/2 = 2; indeed 2 * 1 = 2. With P = −6, the quadratic is 2x^2 − 6x + 4 = 0 = 2(x^2 − 3x + 2) whose roots are 1 and 2. Why Other Options Are Wrong: Only (1, −6) matches both the coefficient value and the other root consistent with Vieta’s relations. Common Pitfalls: Forgetting to divide by the leading coefficient when using Vieta’s formulas or sign mistakes when computing −P/2. Final Answer: 1, −6

Question 19

Solve for roots in terms of parameters: Find the roots of 3a^2 x^2 − a b x − 2 b^2 = 0 in terms of a and b (assume a ≠ 0).

Accepted Answer

Introduction / Context: This is a parameterized quadratic. You are asked to express the two roots in terms of a and b. The quadratic formula applies directly; careful algebraic simplification of the discriminant reveals simple rational multiples of b/a. Given Data / Assumptions: Equation: 3a^2 x^2 − a b x − 2 b^2 = 0 Assume a ≠ 0 to avoid division by zero. Concept / Approach: Use the quadratic formula x = [ab ± √(a^2 b^2 + 24 a^2 b^2)]/(2*3a^2). Factor a^2 b^2 out of the square root to simplify. Then reduce the fraction to a clean form in terms of b/a. Step-by-Step Solution: Discriminant Δ = (−ab)^2 − 4*(3a^2)*(−2b^2) = a^2 b^2 + 24 a^2 b^2 = 25 a^2 b^2 √Δ = 5 a b x = [ab ± 5ab] / (6a^2) = (ab/(6a^2)) * (1 ± 5) = (b/(6a)) * (1 ± 5) Roots: x1 = (b/(6a))*6 = b/a, x2 = (b/(6a))*(−4) = −2b/(3a) Verification / Alternative check: Sum of roots = (b/a) + (−2b/(3a)) = b/(3a) = −(coefficient of x)/(coefficient of x^2) = −(−ab)/(3a^2) = b/(3a). Product = (b/a)*(−2b/(3a)) = −2b^2/(3a^2) = constant term / leading coefficient = (−2b^2)/(3a^2). Checks out. Why Other Options Are Wrong: Sign errors or swapped signs do not satisfy Vieta’s relations for this quadratic. Common Pitfalls: Missing the negative sign in the constant term or mishandling √(25 a^2 b^2) = 5ab can lead to incorrect roots. Final Answer: b/a and −2b/(3a)

Question 20

Identify roots from coefficients: For ax^2 + (4a^2 − 3b)x − 12ab = 0 (a ≠ 0), choose the correct pair of roots.

Accepted Answer

Introduction / Context: Instead of applying the quadratic formula, you can test candidate root pairs with Vieta’s relations. The sum of roots equals −B/A and the product equals C/A. Matching both conditions uniquely determines the correct pair and avoids lengthy algebra. Given Data / Assumptions: Quadratic: ax^2 + (4a^2 − 3b)x − 12ab = 0 Leading coefficient a ≠ 0. Concept / Approach: Let roots be r1 and r2. Then r1 + r2 = −(4a^2 − 3b)/a = −4a + 3b/a and r1*r2 = (−12ab)/a = −12b. Test the options to see which pair has the correct sum and product simultaneously. Step-by-Step Solution: Option C proposes r1 = −4a and r2 = 3b/a. Sum = −4a + 3b/a which matches −(4a^2 − 3b)/a. Product = (−4a)*(3b/a) = −12b which matches C/A. Therefore, Option C satisfies both Vieta conditions exactly. Verification / Alternative check: Other options either flip a sign in the sum or product. Checking them quickly shows inconsistency with at least one of Vieta’s relations. Why Other Options Are Wrong: Wrong sign on either 4a or 3b/a changes the sum and/or product, violating the equation’s coefficients. Common Pitfalls: Forgetting the minus in the constant term or miscomputing −B/A when B = 4a^2 − 3b. Final Answer: −4a and 3b/a

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