Question: Directions : In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer. [NICL (AO) 2014]
I. $ 22x^{2}-x-6=0 $
II. $ 63y^{2}-11y-40=0 $
Options:
A) If $ x\le y $
B) lf $ x<y $
C) If $ x>y $
D) If relationship between x and y cannot be established
E) lf $ x\ge y $
Show Answer
Answer:
Correct Answer: D
Solution:
- I. $ 22x^{2}-x-6=0 $
By splitting the middle term,
$ 22x^{2}+11x-12x-6=0 $
$ 11x,(2x+1)-6,(2x+1)=0 $
$ (11x-6)(2x+1)=0 $
$ x=\frac{6}{11}, $ $ x=-\frac{1}{2} $
value of $ x=( \frac{6}{11},,-\frac{1}{2} ) $
II. $ 63y^{2}-11y-40=0 $
By splitting the middle term.
$ 63y^{2}+45y-56y-40=0 $
$ 9y,(7y+5)-8,(7y+5)=0 $
$ (9y-8)(7y+5)=0 $
$ y=\frac{8}{9}, $ $ y=-\frac{5}{7} $
value of $ y=( \frac{8}{9},-\frac{5}{7} ) $
Hence, relationship between x and y cannot be established.