Quadratic Equations

samarth0711 · Joined: 05 Oct 2005 Posts: 4

Hi All,
I am preparing all alone and so need help even with a few solved ones from Arun sharma.
So please help !!

First doubt -
Q If the roots of eqn ax^2-bx+c=0 differ by 2 then which of the following are true

options are a few expressions in a,b and c

Solution provided in the book -
Let the roots be z and z+2
then z+z+2 = b and z(z+2)=c
simplify these equations to reach one of the options

NOW SHOULD NOT z+z+2 = b/a and z(z+2) = c/a ????????/
Looks like a misprint in the book - ISNT IT ??????

Second doubt -
Q If p and q are the roots of eqn x^2+px+q=0 then what is the value of p

Soln in the book -
As p is a root therefore
p^2+p^2+q=0
2p^2+q=0 ------------------(1)

also as q is a root therefore -
q^2+pq+q=0
q(q+p+1)=0
Therefore
q=0 -------------------------(2)
or
q=-p-1 ----------------------(3)

putting (2) in (1) we get p=0
and
putting (3) in (1) we get p=1 or -1/2

There is nothing wrong with this solution however I am confused what am I doing wrong ?
What I am doing is that as p and q are roots therefore by sum of roots and product of roots formulae should it be not
p+q = -p
and
pq = q

however these equations hold only for p=1
why dont these hold for other two values of p as calculated above ??
Please help !!

nirav_21 · Joined: 10 Oct 2005 Posts: 5 Location: Bhavnagar,Gujarat

Q If the roots of eqn ax^2-bx+c=0 differ by 2 then which of the following are true
Hi samarth,
I am also working alone, and i am happy to know that u are working for it.
Here is your solution:
The roots of the equation differ by 2.
Remember if the roots differ by some value say p and q are the rroots, then p-q=2, remember the following formulae, as such they are derived from the usual equations,
sum of the roots p+q=-b/a
Difference between the roots p-q = sqrt of Delta/a
product of roots pq = c/a
now our question is p-q=2,
ya now i found that there is surely a misprint in the answer,and u r right.

Your second answer: -
Q If p and q are the roots of eqn x^2+px+q=0 then what is the value of p

by the above rules, p+q=-p or 2p = q.----------------------------(i)
pq = q----------------------------------------(ii)
or pq-q=0 or q(p-1)=0 or q=0 or p=1.
if q = 0, then by 2q=p, p=0 forming equal roots 0,0
if p=1, q=1/2.
also p-q = sqrt delta/1
delta = p^2-4q . so p-q =sqrt(p^2-4q)
or (p-q)^2 = p^2 - 4q
or p^2 - 2pq +q^2 = p^2 - 4q.
or q^2 -2pq +4q = 0 ------------------------(iii)
From (i) and (iii), q^2 - q^2 + 4q =0 => q=0, so p=0, again same roots.
From (ii) and (iii), q^2 - q + 4q =0 =>q^2 + 3q=0
=>q(q+3) = 0.
i.e. q=0, or q=-3, if q=0, p=0
if q=-3, p=1.
So the possible outcomes are (p,q) =(0,0) or (1,1/2) or (1,-3).
I havent read your answer,but i have just answered in my method. I hope it would have clarified your doubts.
Hope for a pleasant reply, if u r satisfied.
Bye.

samarth0711 · Joined: 05 Oct 2005 Posts: 4

Thanks a lot for your reply !!
I had lost all hope of getting one and hence just chanced upon checking the forum today to be pleasantly surprised to see your reply.

It will help me a lot as I know now that Arun Sharma - My Quant book, can have misprints and this consequently will help me focus on the time worthiness of the effort when getting stuck.

Thanks again -
and - sorry if it looks cliched and stupid.................still.............Lets meet soon at IIMA !!!!! Smile

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