Equations And Inequations Ques 26
I. $x^{2}+x-12=0$
II. $y^{2}-3 y-10=0$
Directions : In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer If
(1) $x<y$
(2) $x>y$
(3) $x \leq y$
(4) $x \geq y$
(5) $\quad x=y$ or the relationship cannot be established
(United Bank of India PGDBF Manipal Exam,07.08.2016)
Show Answer
Correct Answer: 26.(5)
Solution: (5) I. $x^{2}+x-12=0$
$\Rightarrow x^{2}+4 x-3 x-12=0$
$\Rightarrow x(x+4)-3(x+4)=0$
$\Rightarrow(x-3)(x+4)=0$
$\Rightarrow x=3$ or -4
II. $y^{2}-3 y-10=0$
$\Rightarrow y^{2}-5 y+2 y-10=0$
$\Rightarrow y(y-5)+2(y-5)=0$
$\Rightarrow(y+2)(y-5)=0$
$\Rightarrow y=-2$ or 5
BETA
