In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option.
Give answer :
a: If x > y
b: If x < y
c: If x ≥ y
d: If x ≤ y
e: If x = y or the relationship cannot be established
I. $$15x^{2} + 8x + 1 =0$$
II. $$3y^{2} + 14y + 8 = 0$$
Root of quadratic equation $$(ax^{2} + bx + c =0) = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
For equation I : a=15 , b = 8 and c= 1
On solving the quadratic equation I we get x = (-10/30) or (-6/30)
For equation II : a=3 , b = 14 and c= 8
On solving the quadratic equation I we get y = (-4/6) or (-4)
It is clear that y<x.
Option A is correct answer.
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