In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer If
a:x > y
b: x ≥ y
c: x < y
d: x ≤ y
e: x = y or the relationship cannot be established
$$I.6x^{2} + 19x +15 = 0$$
$$II. 24y^{2} + 11y + 1=0$$
I. 6$$x^{2}$$+19x+15=0
6$$x^{2}$$+10x+9x+15=0
2x(3x+5)+3(3x+5)=0
(2x+3)(3x+5)=0
x=(-3/2) or (-5/3)
II. 24$$y^{2}$$+11y+1=0
24$$y^{2}$$++8y+3y+1=0
8y(3y+1)+1(3y+1)=0
y=(-1/8) or (-1/3)
clearly, x<y
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