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. $$x^{2} + 12x + 35 = 0$$
II. $$7y^{2} + 32y + 16 = 0$$
Root of quadratic equation (ax$$^2$$ + bx + c =0) = $$\frac{-b \pm \sqrt{(b^2 -4ac)}}{2a}$$
For equation I : a=1 , b =12 and c= 35
On solving the quadratic equation I we get x = (-5) or (-6)
For equation II : a=7 , b =32 and c= 16
On solving the quadratic equation I we get y = (-4) or (-8/14)
It is clear that x<y.
Option B is correct answer.
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