Algebra

Equation I: x² + 2x - 8 = 0, so x = (-2 ± √(4 + 32))/2 = (-2 ± 6)/2 = 2 or -4. Equation II: (y-1)² = 0, so y = 1. When x=2, x>y; when x=-4, x

Solving equation (I): x^2 - 16x + 63 = 0 factors as (x-7)(x-9) = 0, giving x = 7 or 9. Solving equation (II): y^2 - 21y + 110 = 0 factors as (y-10)(y-11) = 0, giving y = 10 or 11. Comparing all possible values, both solutions for x (7 and 9) are less than both solutions for y (10 and 11), so x < y always holds.

Solving equation (I): x^2 - 13x + 40 = 0 factors as (x-5)(x-8) = 0, giving x = 5 or 8. Solving equation (II): y^2 - 10y + 21 = 0 factors as (y-3)(y-7) = 0, giving y = 3 or 7. When x=5, y could be 3 (x>y) or 7 (xy). Since the relationship varies depending on which values we pick, it cannot be uniquely determined.

Solving equation (I): x^2 - 18x + 77 = 0 factors as (x-7)(x-11) = 0, giving x = 7 or 11. Solving equation (II): y^2 - 24y + 143 = 0 factors as (y-11)(y-13) = 0, giving y = 11 or 13. When x=7, y can be 11 or 13 (x

For x² - x - 1 = 0, sum a + ß = 1 and product aß = -1. First find a² + ß² = (a + ß)² - 2aß = 1 + 2 = 3. Then a⁴ + ß⁴ = (a² + ß²)² - 2(aß)² = 9 - 2 = 7.

For a quadratic x² + bx + c = 0, sum of roots = -b and product = c. Given sum = 9, we have -b = 9, so b = -9. Given product = 9, we have c = 9. So the equation is x² - 9x + 9 = 0.

Solve (I): 6x² - x - 2 = 0 factors to (3x - 2)(2x + 1) = 0, giving x = 2/3 or x = -1/2. Solve (II): 5y² - 18y + 9 = 0. Discriminant = 324 - 180 = 144. y = (18 ± 12)/10 = 3 or 0.6. Comparing: x = 2/3 ≈ 0.67 and y = 0.6 means x > y for this pair. But x = -1/2 and y = 3 means x < y. Since x can be greater or less than y, relationship cannot be established.

Solve (I): 5x² - 13x - 6 = 0 factors to (5x + 2)(x - 3) = 0, giving x = 3 or x = -0.4. Solve (II): 3y² + 8y + 4 = 0 factors to (3y + 2)(y + 2) = 0, giving y = -2 or y = -2/3. Comparing: x = 3 > both y values; x = -0.4 > -2 and > -0.67. So x > y always.

Solve (I): 3x² - 4x - 4 = 0 factors to (3x + 2)(x - 2) = 0, giving x = 2 or x = -2/3. Solve (II): 3y² + 14y + 8 = 0 factors to (3y + 2)(y + 4) = 0, giving y = -2/3 or y = -4. Comparing: x = 2 > both y values (x ≥ y true); x = -2/3 = -2/3 (x ≥ y true); x = -2/3 > -4 (x ≥ y true). So x ≥ y always.

Solve (I): 2x² + 17x + 30 = 0 factors to (2x + 5)(x + 6) = 0, giving x = -2.5 or x = -6. Solve (II): 6y² - 5y - 6 = 0 factors to (3y - 4)(2y + 3) = 0, giving y = 4/3 or y = -1.5. Comparing: x = -2.5 < y = -1.5 (x < y true); x = -2.5 < y = 4/3 (x < y true); x = -6 < both y values. So x < y always.

Equation I: x² - 21x + 68 = 0 factors to (x-17)(x-4)=0, so x = 17 or x = 4. Equation II: y² - 3y - 4 = 0 factors to (y-4)(y+1)=0, so y = 4 or y = -1. When x=17, y can be 4 or -1, so x > y. When x=4, y can be 4 (x=y) or -1 (x > y). Therefore x ≥ y in all cases. Option D is correct.

Solving the system: 3x + 5y = 29 and 4x - y = 8. From the second equation, y = 4x - 8. Substituting: 3x + 5(4x-8) = 29, giving 3x + 20x - 40 = 29, so 23x = 69 and x = 3. Then y = 4(3) - 8 = 4. Since x = 3 and y = 4, we have x < y. Option C is correct.

Solve equation (I): x^2 - 10x + 24 = 0 factors as (x-4)(x-6) = 0, giving x = 4 or 6. Solve equation (II): 2y^2 - 12y + 18 = 0 simplifies to y^2 - 6y + 9 = 0, which factors as (y-3)^2 = 0, so y = 3 (repeated root). Since both values of x (4 and 6) are greater than y (3), the relationship is x > y.

For x² - 6x + 6 = 0 with roots α and β: α+β = 6, αβ = 6. α²+β² = (α+β)² - 2αβ = 36 - 12 = 24. α³+β³ = (α+β)³ - 3αβ(α+β) = 216 - 3(6)(6) = 216 - 108 = 108. Total = α³+β³ + α²+β² + α+β = 108 + 24 + 6 = 138.

For m²-19m=-88: m²-19m+88=0, (m-8)(m-11)=0, so m=8 or 11. For n²-48n+576=0: (n-24)²=0, so n=24 (repeated root). Both possible m values (8, 11) are less than n=24, so m