Quadratic Equations

For equation (I), factoring gives x^2 - 102x + 2240 = (x - 70)(x - 32) = 0, so x = 70 or x = 32. For equation (II), y^2 - 156y + 2240 = (y - 140)(y - 16) = 0, so y = 140 or y = 16. When x = 70, it can be greater than y = 16 but less than y = 140. When x = 32, it can be greater than y = 16 but less than y = 140. Since x can be greater than, less than, or even equal to y depending on which root you pick, no definitive relationship exists.

For equation (I), using the quadratic formula: x = [11√3 ± √((11√3)^2 - 4×84)]/2 = [11√3 ± √(363 - 336)]/2 = [11√3 ± √27]/2 = [11√3 ± 3√3]/2, giving x = 7√3 or x = 4√3. For equation (II), the variable should be y, not x (this is a typo in the question). Solving y^2 - 16√3y + 84 = 0 gives y = [16√3 ± √768]/2 = [16√3 ± 8√3]/2, giving y = 14√3 or y = 6√3. Numerically: x ≈ 12.12 or 6.93, y ≈ 24.25 or 10.39. Depending on the pair chosen, x can be less than or greater than y, so no relationship can be established.

Solving Quantity I: x^2 - 12x + 36 = 0 factors to (x-6)(x-6) = 0, giving x = 6. Solving Quantity II: y^2 - 18y + 81 = 0 factors to (y-9)(y-9) = 0, giving y = 9. Comparing: 6 < 9, so Q1 < Q2. The claimed answer B is correct.

Equation I: x²-11x+30=0 gives x=5,6 (roots of 5 and 6). Equation II: y²-9y+20=0 gives y=4,5 (roots of 4 and 5). Comparing all combinations: when x=5,y=5 (x=y); when x=6,y=4 (x>y); when x=6,y=5 (x>y); when x=5,y=4 (x>y). In all cases x ≥ y, making option C correct.

Solving the first equation: x^2 - 30x + 221 = 0 has discriminant 16, giving roots x = 13 and x = 17. Solving the second: y^2 - 34y + 285 = 0 has discriminant 16, giving roots y = 15 and y = 19. Comparing all possible values, x can be 13 or 17 while y can be 15 or 19. When x=17 and y=15, we have x>y, but when x=13 and y=19, we have x

For x² - 7x + 12 = 0: factors as (x-3)(x-4) = 0, so x = 3 or 4. For y² - 11y + 30 = 0: factors as (y-5)(y-6) = 0, so y = 5 or 6. Since all possible values of x (3, 4) are less than all possible values of y (5, 6), we can conclude x < y is always true.

For 2x² + 10x - 28 = 0: dividing by 2 gives x² + 5x - 14 = 0, which factors as (x+7)(x-2) = 0, so x = -7 or 2. For 4y² - 19y + 21 = 0: using quadratic formula gives y = (19 ± √361-336)/8 = (19 ± 5)/8, so y = 3 or 1.75. Comparing: x can be -7 or 2, y can be 1.75 or 3. The values overlap (2 < 3 but 2 > 1.75), so no clear relationship exists.

For 12x² + 37x + 21 = 0: factors as (3x+7)(4x+3) = 0, giving x = -7/3 ≈ -2.33 or x = -3/4 = -0.75. For 6y² - 32y + 42 = 0: dividing by 2 gives 3y² - 16y + 21 = 0, which factors as (y-3)(3y-7) = 0, so y = 3 or y = 7/3 ≈ 2.33. All x values (-2.33, -0.75) are less than all y values (2.33, 3), so x < y.

For 24x² - 122x + 10 = 0: using quadratic formula, x = (122 ± √14884 - 960)/48 = (122 ± √13924)/48. √13924 ≈ 118, so x ≈ (122 + 118)/48 ≈ 5 or x ≈ (122 - 118)/48 ≈ 0.08. For 2y² + 5y + 3 = 0: discriminant = 25 - 24 = 1, so y = (-5 ± 1)/4, giving y = -1 or y = -1.5. Since all x values (0.08, 5) are greater than all y values (-1.5, -1), we have x > y.

For 14x²+25x+9=0: factors are (2x+1)(7x+9), giving x = -1/2 or x = -9/7. For 2y²-y-1=0: factors are (2y+1)(y-1), giving y = -1/2 or y = 1. Comparing values: x = -9/7 ≈ -1.29, x = -0.5; y = -0.5, y = 1. All x values are less than or equal to all y values, so x ≤ y is true (x equals y when both are -0.5, and x is less than y in other cases).

Solve 6x^2 + x - 1 = 0: x = [-1 ± √(1 + 24)]/12 = (-1 ± 5)/12, so x = 1/3 or x = -1/2. Solve 6y^2 + 7y - 3 = 0: y = [-7 ± √(49 + 72)]/12 = (-7 ± 11)/12, so y = 1/3 or y = -3/2. Since x = 1/3 matches y = 1/3, but x = -1/2 could be greater or less than y depending on which y value is chosen, multiple relationships are possible. When both roots are 1/3, x = y. When x = -1/2 and y = 1/3, x < y. When x = -1/2 and y = -3/2, x > y. Therefore, no single relationship always holds.

Solve 12x^2 + 28x + 15 = 0: x = [-28 ± √(784 - 720)]/24 = (-28 ± 8)/24, so x = -20/24 = -5/6 or x = -36/24 = -3/2. Solve 6y^2 + y - 2 = 0: y = [-1 ± √(1 + 48)]/12 = (-1 ± 7)/12, so y = 6/12 = 1/2 or y = -8/12 = -2/3. Compare: x = -5/6 with y = 1/2 gives x < y (-0.833 < 0.5). x = -5/6 with y = -2/3 gives x < y (-0.833 < -0.667). x = -3/2 with y = 1/2 gives x < y (-1.5 < 0.5). x = -3/2 with y = -2/3 gives x < y (-1.5 < -0.667). In all four cases, x < y.

Solve 6x^2 + 9x + 3 = 0: Divide by 3: 2x^2 + 3x + 1 = 0. Factor: (2x + 1)(x + 1) = 0, so x = -1/2 or x = -1. Solve 4y^2 + 6y + 2 = 0: Divide by 2: 2y^2 + 3y + 1 = 0. This is identical to the simplified x equation, so y = -1/2 or y = -1. Compare: x = -1/2 vs y = -1/2 gives x = y. x = -1 vs y = -1 gives x = y. x = -1/2 vs y = -1 gives x > y (-0.5 > -1). x = -1 vs y = -1/2 gives x < y (-1 < -0.5). Since both equality and inequality occur depending on which roots are compared, the relationship cannot be established.

Equation (I): x² = 324 gives x = ±18 (positive and negative roots). Equation (II): y = √324 = 18 (principal square root is positive). When x = 18, x = y. When x = -18, x < y. In both cases, x ≤ y is true. Option D correctly captures this relationship.

Solve equation (I): x² - 4x - 12 = 0 gives x = -2 or x = 6. Solve equation (II): 3y² - 8y + 5 = 0 gives y = 1 or y = 5/3. Comparing values: when x = -2, x < y (since both y values are positive); when x = 6, x > y (since both y values are less than 6). Since the relationship changes depending on which roots we compare, no consistent relationship can be established.