quadratic equation and its solution

Aug 2011
143
1
Hi,

The following question I encountered in the National Talent Search Examination 2019 in India (State - Andra Pradesh)

If -2 is a root of the quadratic equation x^2 –px + 6 = 0 and x^2 + px – k = 0 has equal roots, then the value of k is

(1) 6 (2) 10 (3) 14 (4) 18

They have given the answer as 14.

How to do? With the normal understanding of root and equal roots I am not able to get any of the choices as my answer.

Help solicited.

With regards

Aranga
 
Jan 2009
448
127
It seems like you would need to leverage the root coefficient relationship and set up a system of equations to solve for k.

Currently at work, so if this doesn't help, I can try to take a look after work.

EDIT://
You can also use the fact that -2 is a root to solve for p in terms of k. ie. x^2 - px + 6 = x^2 + px - k
 
Last edited:

Debsta

MHF Helper
Oct 2009
1,343
620
Brisbane
I get k = -25/4. Is that what you got?

If -2 is a root of the quadratic equation x^2 –px + 6 = 0 then \(\displaystyle (-2)^2+2p+6=0 => p=-5\)

If x^2 + px – k = 0 has equal roots then \(\displaystyle \Delta = p^2+4k = 0 => (-5)^2+4k=0 => k=\frac{-25}{4}\)
 
Last edited:

Debsta

MHF Helper
Oct 2009
1,343
620
Brisbane
If k=14, that would mean that \(\displaystyle x^2 + px -14 =0\) must have equal roots.

To have equal roots \(\displaystyle \Delta= b^2 -4ac=p^2 +56=0\)

This has no real solutions for p.

We are told that one root of \(\displaystyle x^2 -px+6\) is -2, so the second root must be real (ie we don't need to consider complex roots).

Basically, k can't be 14.

I would say there is a misprint in the question if the answer is supposed to be 14. Are you sure you've copied the question exactly as given?
 

Debsta

MHF Helper
Oct 2009
1,343
620
Brisbane
Something is wrong with either the question or the answers.
k=-25/4 as shown before.
 
  • Like
Reactions: arangu1508
Jan 2009
448
127
Thanks Debsta. Ignore the edit part of my answer as it isn't valid.
There is an error in the question when it says both equations have equal roots.

@arangu1508
I just used the root-coefficient relationship as I advised in my first post and got the answer of k=14, but this is without the assumption that all roots of both polynomial equations are the same.

Applying it to the first equation \(\displaystyle x^2 –px + 6 = 0\) I get the following
\(\displaystyle r_1 + r_2 = -b/a = p \)
\(\displaystyle r_1 \cdot r_2 = c/a = 6 \)
With \(\displaystyle r_1 = -2 \) I get \(\displaystyle r_2 = -3\) and \(\displaystyle p = -5\)

I can then apply the root-coefficient relationship again to the second equation, and solve for k, knowing p.
I have given you half the answer, so I want you to try the remaining half on your own.
 
  • Like
Reactions: arangu1508

Debsta

MHF Helper
Oct 2009
1,343
620
Brisbane
Thanks Debsta. Ignore the edit part of my answer as it isn't valid.
There is an error in the question when it says both equations have equal roots.
The question doesn't say that both equations have equal roots. Only the second equation has equal roots, ie one distinct root, ie discriminant equals 0.
 
  • Like
Reactions: arangu1508