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
$$I. 8x^{2} + 26x+ 15 =0$$
$$II. 4y^{2} + 24y+ 35 = 0$$
I. 8$$x^{2}$$+26x+15=0
8$$x^{2}$$+20x+6x+15=0
4x(2x+5)+3(2x+5)=0
(4x+3)(2x+5)=0
x=(-3/4) or (-5/4)
II . 4$$y^{2}$$+24y+35=0
4$$y^{2}$$+10y+14y+35=0
2y(2y+5)+7(2y+5)=0
(2y+7)(2y+5)=0
y=(-7/2) or (-5/2)
Clearly x>y
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