Quadratic Equation creates problem.

Algebra Level pending

If i,j be the roots of the equation x 2 + p x − q = 0 x^{2}+px -q=0 and k,l be the roots of x 2 + p x + r = 0 x^{2}+px+r=0 , (q+r) not be 0 Then [ ( i − k ) × ( i − l ) ] ÷ [ ( j − k ) × ( j − l ) ] i s [(i-k)×(i-l)]÷[(j-k)×(j-l)] is


The answer is 1.

Arghyanil Dey by Arghyanil Dey , 26, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Maharnab Mitra
Apr 29, 2014

Assume p=2, q=0, r=1 and BANG ON! you get the answer..................... I know it seems frustrating to the problem creator but anyway........................ :)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Good , but I want the proper process.

Arghyanil Dey - 7 years, 1 month ago

Actually, I too want to know the proper process!

Maharnab Mitra - 7 years, 1 month ago

An even quicker solution to this Can be that the roots have not been differentiated (in order), which means if the roots to the equation 1 are (2,3) it can be taken both ways i=2,j=3 OR i=3,j=2 and this selection results in different numerator and denominators say N & D. So, the question here is not affected by the chosen numerator or denominator which implies: N/D = D/N
and this is possible only in 2 cases, when N=D OR when N=-D It gives only 2 possible outcomes: -1 and 1 and you are given three chances ;)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...