Equations And Inequations Ques 162
I. $x^{2}-14 x+48=0$
II. $y^{2}+6=5 y$
Directions: In the following questions, two equations I and II have been given. You have to solve both equations and give answer if
(1) $x>y$
(2) $x \geq y$
(3) $x<y$
(4) $x \leq y$
(5) $x=y$ or the relation cannot be established.
(Bank of Baroda PO Exam. 14.08.2014)
Show Answer
Correct Answer: (1)
Solution: (1) I. $x^{2}-14 x+48=0$
$\Rightarrow x^{2}-8 x-6 x+48=0$
$\Rightarrow x(x-8)-6(x-8)=0$
$\Rightarrow(x-6)(x-8)=0$
$\Rightarrow x=6$ or 8
II. $y^{2}-5 y+6=0$
$\Rightarrow y^{2}-3 y-2 y+6=0$
$\Rightarrow y(y-3)-2(y-3)=0$
$\Rightarrow(y-3)(y-2)=0$
$\Rightarrow y=2$ or 3
BETA
