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.$$9x^{2} -37 x + 30 = 0$$
II.$$3y^{2} -19y + 30 = 0$$
Root of quadratic equation $$(ax^{2} + bx + c =0)$$ = $$\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
For equation I : a=9 , b = -37 and c= 30
On solving the quadratic equation I we get x = (3) or (20/18)
For equation II : a=3 , b = -19 and c= 30
On solving the quadratic equation I we get y = (3) or (20/6)
It is clear that y>x
Option A is correct answer.
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