In these questions, two equations numbered I and H are given. You have to solve both the equations and mark the appropriate option.
Give answer :
a: If relationship between x and y cannot be established
b: If x < y
c: If x > y
d: If x ≤ y
e: If x ≥ y
I. $$2x^{2} - 17x + 36 = 0$$
II. $$5y^{2} - 33y + 40 = 0$$
Root of quadratic equation $$(ax^{2} + bx + c =0)$$ = $$\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
For equation I : a=2 , b = -17 and c= 36
On solving the quadratic equation I we get x = (4) or (18/4)
For equation II : a=5 , b = -33 and c= 40
On solving the quadratic equation I we get y = (5) or (3/2)
If y=5 then y>x
If y = 3/2 then y<x.
A relationship cannot be established
Option A is correct answer.
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