Quantity I. 'x': x2 + x - 6 = 0 Quantity II. 'y': y2 + 7y + 12 = 0
-
a) Quantity I > Quantity II
-
b) Quantity I < Quantity II
-
c) Quantity I ≥ Quantity II
-
d) Quantity I ≤ Quantity II
-
e) Quantity I = Quantity II or No relation
Quantity I: x^2 + x - 6 = 0 => (x+3)(x-2) = 0 => x = -3, 2. Quantity II: y^2 + 7y + 12 = 0 => (y+3)(y+4) = 0 => y = -3, -4. Comparing sets {-3, 2} and {-3, -4}: 2 > -3, 2 > -4, -3 = -3, -3 > -4. In all cases, x >= y.
For the first equation x2 + x - 6 = 0, splitting the middle term gives x2 + 3x - 2x - 6 = 0, which simplifies to x(x + 3) - 2(x + 3) = 0; this yields the roots x = -3 or x = 2. For the second equation y2 + 7y + 12 = 0, splitting the middle term gives y2 + 4y + 3y + 12 = 0, which simplifies to (y + 4)(y + 3) = 0; this yields the roots y = -4 or y = -3. When comparing the values, x = 2 is greater than both y values, but x = -3 is equal to y = -3, so the relation is x is greater than or equal to y.