Equations And Inequations Ques 165
I. $x^{2}-28 x+196=0$
II. $y^{2}=196$
Directions : In each of the following questions two equations I and II have been 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 relationship cannot be established between them
(IDBI Officer Grade Exam. 22.08.2014)
Show Answer
Correct Answer: (2)
Solution: (2) I. $x^{2}-28 x+196=0$
$\Rightarrow x^{2}-2 \cdot x \cdot 14+196=0$
$\Rightarrow(x-14)^{2}=0$
$\Rightarrow x-14=0$
$\Rightarrow x=14$
II. $y^{2}=196$
$\therefore y=\sqrt{196}= \pm 14$
Clearly, $x \geq y$
BETA
