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Instructions
In the following questions two equations numbered I and II are given. You have to solve both the equations and —
Give answer a: if × > y
Give answer b: if × ≥ y
Give answer c: if × < y
Give answer d: if × ≤ y
Give answer e: if × = y or the relationship cannot be established.
Solution
I. $$3x^{2} + 14x + 15 = 0$$
=> $$3x^2 + 9x + 5x + 15 = 0$$
=> $$(3x + 5) (x + 3) = 0$$
=> $$x = \frac{-5}{3} , -3$$
II. $$6y^{2} + 17y + 12 = 0$$
=> $$6y^2 + 9y + 8y + 12 = 0$$
=> $$(3y + 4) (2y + 3) = 0$$
=> $$y = \frac{-4}{3} , \frac{-3}{2}$$
$$\therefore y > x$$
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