quadratic equation

- mathmath333

quadratic equation

- chestercat

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- mathmath333

\(\large \color{black}{\begin{align}& \normalsize \text{If}\ p\ \text{and}\ q\ \text{are the roots of the equation} \hspace{.33em}\\~\\
& x^2+px+q=0,\normalsize \text{then find the value of}\ p.\ \hspace{.33em}\\~\\
\end{align}}\)

- mathmath333

\(\large \color{black}{\begin{align}& p+q=-p\hspace{.33em}\\~\\
& pq=p \hspace{.33em}\\~\\
\end{align}}\)
from the sum and product of roots i got this , is this correct.

- anonymous

yes surely...

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## More answers

- anonymous

For equation : \(ax + by + c = 0\),
Sum of Roots = \(\frac{-b}{a}\)
Product of Roots = \(\frac{c}{a}\)

- anonymous

Are you sure p = 1??

- mathmath333

\(p=-1/2\)

- anonymous

If p = 1 then from equation 2, q = 1.. But from equation 1, 2p = -q does not get satisfied.

- anonymous

This looks good.. :)

- mathmath333

lol but the answer given is wrong by book it says the answer is
\(\large \color{black}{\begin{align}p=\{0,1,-\dfrac{-1}{2}\} \hspace{.33em}\\~\\
\end{align}}\)

- anonymous

0 and -1/2 are all right, but how p = 1 is possible?

- mathmath333

how did u find \(0\) as a solution.

- anonymous

Wait, there is a slight mistake that you have done..

- anonymous

And I also being mad, I did not check that.. What you got for Product of Roots?

- anonymous

\[pq = q\]
This must be the product no??

- mathmath333

?

- anonymous

See, the formulae I gave above, product of roots = c/a.. in your question c = q and a = 1..

- Empty

The only way \(pq=p\) can be true is if:
\(q=1\) OR \(p=0\)

- mathmath333

ok how about \(p=1\)

- anonymous

@Empty that should be q on right side, not p..

- mathmath333

oh ur right .

- Empty

?
\(pq=p\) is the same thing as \(p=qp\)

- anonymous

Just solve these equations:
\[p + q = -p \\ pq = q\]

- anonymous

How they are same?

- Empty

Ohh I see, I switched out my p's and q's. Oh well haha, same idea really since they're completely symmetric. XD

- anonymous

@mathmath333 if you use pq = p (the original one you did above), p = 1 as solution is not possible in any case..

- mathmath333

ok i got \(p=0,1\) how about \(p=-1/2\) now ?

- anonymous

That should be pq = q...

- Empty

\(\large \color{black}{\begin{align}& p+q=-p\hspace{.33em}\\~\\
& pq=p \hspace{.33em}\\~\\
\end{align}}\)
What mathmath wrote is wrong here, I went off this, but really the last equation is what @waterineyes is trying to tell us!

- mathmath333

lol big drama

- mathmath333

why \(p=1\) is not possible ?

- Empty

p=1 is possible only when q=0. So you have to work out the consequence of that in the other equation.

- mathmath333

its so confusin

- anonymous

Hey @mathmath333 listen to me, what is product of roots you get?

- mathmath333

\(pq=q\)

- anonymous

yea that is great..

- anonymous

Now, you get the cases of 0 and 1 ??

- mathmath333

yes i do

- anonymous

So, we left with how p = -/12 right?

- mathmath333

yes

- anonymous

*-1/2..

- anonymous

Put p = -1/2 in first equation and find q...

- mathmath333

\(q=1\)

- anonymous

Good...

- anonymous

But it is not satisfying the other equation.. :P

- mathmath333

how ?

- anonymous

What you get when you do p times q ?? -1/2 times 1 = -1/2 but on other side you have q which is 1 and -1/2 not equal to 1..

- anonymous

\[pq = q \\ \frac{-1}{2} \times 1 = 1 \\ \frac{-1}{2} \ne 1\]

- anonymous

@Empty if you have something to correct or say, then you can proceed, I have tried my bit. :)

- mathmath333

wait

- anonymous

Yeah waiting, ordering coffee or tea something? :P

- mathmath333

is \(0\) and \(1\) both correct

- anonymous

@SolomonZelman your help is needed here..

- mathmath333

he came first but left

- anonymous

Yes 0 and 1 both are satisfying but -1/2 is not..

- mathmath333

seee this wolfram link
http://www.wolframalpha.com/input/?i=solve+2p%5E2%2Bq%3D0%2Cq%5E2%2Bpq%2Bq%3D0+over+reals

- mathmath333

- anonymous

Now wolfram will tell the secrets of us that we don't know anything. :P

- mathmath333

my head hurts

- mathmath333

- anonymous

And I call the first master : Please save us : @mukushla ..

- freckles

\[p+q=-p \\ pq=q \]
these are the equations agreed?

- mathmath333

yes

- freckles

\[\text{ the second equation we have } \\ pq=q \\ pq-q=0 \\ q(p-1) =0 \implies q=0 \text{ or } p=1 \\ \text{ going back to first equation } \\ \text{ if } q=0 \text{ we have } p=-p \text{ which forces } p=0 \\ \text{ if } p=1 \text{ then } 1+q=-1 \implies q=-2 \\ \text{ so far we have the following pairs of solutions } \\ (p,q) \\ (0,0) \\ (1,-2)\]

- freckles

I think though you say we have another solution

- freckles

so let's see if we can play with the equations more to find it

- mathmath333

i thhink that is same as of frekles

- freckles

\[p=\frac{-1}{2} \text{ trying first } \\ \frac{-1}{2}+q=-(\frac{-1}{2}) \\ \frac{-1}{2}+q=\frac{1}{2} \\ q=1 \\\text{ so if } p=\frac{-1}{2} \text{ is a solution } \\ \text{ then we should get } q=1 \text{ for the second equation too} \\ \frac{-1}{2}q=q \\ \frac{-1}{2}q-q=0 \\ q=0 \\ \text{ so } p=\frac{-1}{2} \text{ doesn't seem to work out }\]

- freckles

the only solutions I think there to be is (0,0) and (1,-2)
where those ordered pairs are in the form (p,q)

- mathmath333

but wolfram and my book does say \(p=-1/2\) also.

- freckles

can i see the link to wolfram

- freckles

because my wolfram checks my solutions just fine

- freckles

http://www.wolframalpha.com/input/?i=p%2Bq%3D-p+%2C+pq%3Dq

- mathmath333

http://www.wolframalpha.com/input/?i=solve+2p%5E2%2Bq%3D0%2Cq%5E2%2Bpq%2Bq%3D0+over+reals

- freckles

ok I see what you did:
you pluggged p and q into the quadratic
and didn't use veita's formula to find p,q for the -1/2,-1/2 solution
hmm... so let's see
you have p=-1/2 and q=-1/2
let's see if I can solve that other system by hand ..

- anonymous

If you use both p and q values, then also I think both equations will not get satisfied..

- mathmath333

but vieta's formula shouls work too, something is missing

- freckles

\[2p^2+q=0 \\ q^2+pq+q=0 \\ 2p^2=-q \\ -2p^2=q \\ (-2p^2)^2+p(-2p^2)+(-2p^2)=0 \\ 4p^4-2p^3-2p^2=0 \\ 2p^2(2p^2-p-1)=0 \\ p=0 \text{ or } 2p^2-p-1=0 \\ p=0 \text{ or } (2p+1)(p-1)=0 \\ p=0 \text{ or } p=\frac{-1}{2} \text{ or } p=1 \]
hmm... I don't know why vieta's formula isn't working for that one p

- mathmath333

:D

- freckles

but I would like to know why it isn't working for veita's formula :(

- mathmath333

thats a sign of good mathmatician.

- freckles

http://www.wolframalpha.com/input/?i=x%5E2-1%2F2x-1%2F2%3D0
since p=-1/2 then q=-2p^2=-2(-1/2)^2=-2(-1/4)=-1/2
so we have p=-1/2 and q=-1/2
but that wolfram link doesn't give the roots -1/2 and -1/2
if gives the roots -1/2 and 1

- freckles

so I think p=-1/2 shouldn't be a solution

- mathmath333

:(

- freckles

so veita's formula I think does work
and the other way gave too many solutions (solutions that needed to be checked)

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