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. $$6x^{2} - 25x + 14 =0$$
II. $$9y^{2} -9y + 2 =0$$
Root of quadratic equation $$(ax^{2} + bx + c =0)$$ = $$\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
For equation I : a=6 , b = -25 and c= 14
On solving the quadratic equation I we get x = (8/12) or (42/12)
For equation II : a=9 , b = -9 and c= 2
On solving the quadratic equation I we get y = (12/18) or (6/18)
It is clear that x ≥ y
Option C is correct answer.
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