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In each of the following questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.Give answer Ifa: x > yb: x ? yc: x ? yd: x < ye: Relationship between x and y cannot be established

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In each of the following questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.
Give answer If
a: x > y
b: x ? y
c: x ? y
d: x < y
e: Relationship between x and y cannot be established


I.$$x^{2}-x-12= 0$$
II. $$y^{2}+4y+4=0$$
1). x > y
2). x ≤ y
3). x ≥ y
4). x < y

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1 answers

2 vote
Answered by on | Votes 2 |
Solution

I.$x^{2} - x - 12 = 0$

=> $x^2 - 4x + 3x - 12 = 0$

=> $x (x - 4) + 3 (x - 4) = 0$

=> $(x - 4) (x + 3) = 0$

=> $x = 4 , -3$

II.$y^{2} + 4y + 4 = 0$

=> $y^2 + 2y + 2y + 4 = 0$

=> $y (y + 2) + 2 (y + 2) = 0$

=> $(y + 2) (y + 2) = 0$

=> $y = -2 , -2$

Because $4 > -2$ and $-2 > -3$

$\therefore$ No relation can be established.

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