Equations And Inequations Ques 137
I. $16 x^{2}+20 x+6=0$
II. $10 y^{2}+38 y+24=0$
Directions : In each of these questions, two equations are given. You have to solve these equations and find out the values of $x$ and $y$ and Give answer If
(1) $x<y$
(2) $x>y$
(3) $x \leq y$
(4) $x \geq y$
(5) $x=y$
(Punjab & Sind Bank PO Exam. 23.01.2011)
Show Answer
Correct Answer: (2)
Solution: (2) I. $16 x^{2}+20 x+6=0$
$\Rightarrow 8 x^{2}+10 x+3=0$
$\Rightarrow 8 x^{2}+6 x+4 x+3=0$
$\Rightarrow 2 x(4 x+3)+1(4 x+3)=0$
$\Rightarrow(2 x+1)(4 x+3)=0$
$\therefore x=-\frac{1}{2}$ or $-\frac{3}{4}$
II. $10 y^{2}+38 y+24=0$
$\Rightarrow 5 y^{2}+19 y+12=0$
$\Rightarrow 5 y^{2}+15 y+4 y+12=0$
$\Rightarrow 5 y(y+3)+4(y+3)=0$
$\Rightarrow(y+3)(5 y+4)=0$
$\therefore y=-3$ or $-\frac{4}{5}$
Clearly, $x>y$
BETA
