Equations And Inequations Ques 76
I. $x^{2}+9 x+20=0$
II. $y^{2}=16$
Directions : In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer
(1) if $x>y$
(2) if $x \geq y$
(3) if $x<y$
(4) if $x \leq y$
(5) if $x=y$ or the relationship between $x$ and $y$ cannot be established.
Show Answer
Correct Answer: 76.(4)
Solution: (4) I. $x^{2}+9 x+20=0$
$\Rightarrow x^{2}+4 x+5 x+20=0$
$\Rightarrow x(x+4)+5(x+4)=0$
$\Rightarrow(x+4)(x+5)=0$
$\Rightarrow x=-4$ or, -5
II. $y^{2}=16$
$\Rightarrow y=\sqrt{16}= \pm 4$
Clearly, $x \leq y$
BETA
