mathmath333
  • mathmath333
quadratic equation
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • 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
  • 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
  • anonymous
yes surely...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
For equation : \(ax + by + c = 0\), Sum of Roots = \(\frac{-b}{a}\) Product of Roots = \(\frac{c}{a}\)
anonymous
  • anonymous
Are you sure p = 1??
mathmath333
  • mathmath333
\(p=-1/2\)
anonymous
  • anonymous
If p = 1 then from equation 2, q = 1.. But from equation 1, 2p = -q does not get satisfied.
anonymous
  • anonymous
This looks good.. :)
mathmath333
  • 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
  • anonymous
0 and -1/2 are all right, but how p = 1 is possible?
mathmath333
  • mathmath333
how did u find \(0\) as a solution.
anonymous
  • anonymous
Wait, there is a slight mistake that you have done..
anonymous
  • anonymous
And I also being mad, I did not check that.. What you got for Product of Roots?
anonymous
  • anonymous
\[pq = q\] This must be the product no??
mathmath333
  • mathmath333
?
anonymous
  • anonymous
See, the formulae I gave above, product of roots = c/a.. in your question c = q and a = 1..
Empty
  • Empty
The only way \(pq=p\) can be true is if: \(q=1\) OR \(p=0\)
mathmath333
  • mathmath333
ok how about \(p=1\)
anonymous
  • anonymous
@Empty that should be q on right side, not p..
mathmath333
  • mathmath333
oh ur right .
Empty
  • Empty
? \(pq=p\) is the same thing as \(p=qp\)
anonymous
  • anonymous
Just solve these equations: \[p + q = -p \\ pq = q\]
anonymous
  • anonymous
How they are same?
Empty
  • 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
  • anonymous
@mathmath333 if you use pq = p (the original one you did above), p = 1 as solution is not possible in any case..
mathmath333
  • mathmath333
ok i got \(p=0,1\) how about \(p=-1/2\) now ?
anonymous
  • anonymous
That should be pq = q...
Empty
  • 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
  • mathmath333
lol big drama
mathmath333
  • mathmath333
why \(p=1\) is not possible ?
Empty
  • Empty
p=1 is possible only when q=0. So you have to work out the consequence of that in the other equation.
mathmath333
  • mathmath333
its so confusin
anonymous
  • anonymous
Hey @mathmath333 listen to me, what is product of roots you get?
mathmath333
  • mathmath333
\(pq=q\)
anonymous
  • anonymous
yea that is great..
anonymous
  • anonymous
Now, you get the cases of 0 and 1 ??
mathmath333
  • mathmath333
yes i do
anonymous
  • anonymous
So, we left with how p = -/12 right?
mathmath333
  • mathmath333
yes
anonymous
  • anonymous
*-1/2..
anonymous
  • anonymous
Put p = -1/2 in first equation and find q...
mathmath333
  • mathmath333
\(q=1\)
anonymous
  • anonymous
Good...
anonymous
  • anonymous
But it is not satisfying the other equation.. :P
mathmath333
  • mathmath333
how ?
anonymous
  • 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
  • anonymous
\[pq = q \\ \frac{-1}{2} \times 1 = 1 \\ \frac{-1}{2} \ne 1\]
anonymous
  • anonymous
@Empty if you have something to correct or say, then you can proceed, I have tried my bit. :)
mathmath333
  • mathmath333
wait
anonymous
  • anonymous
Yeah waiting, ordering coffee or tea something? :P
mathmath333
  • mathmath333
is \(0\) and \(1\) both correct
anonymous
  • anonymous
@SolomonZelman your help is needed here..
mathmath333
  • mathmath333
he came first but left
anonymous
  • anonymous
Yes 0 and 1 both are satisfying but -1/2 is not..
mathmath333
  • mathmath333
seee this wolfram link http://www.wolframalpha.com/input/?i=solve+2p%5E2%2Bq%3D0%2Cq%5E2%2Bpq%2Bq%3D0+over+reals
mathmath333
  • mathmath333
@waterineyes
anonymous
  • anonymous
Now wolfram will tell the secrets of us that we don't know anything. :P
mathmath333
  • mathmath333
my head hurts
mathmath333
  • mathmath333
@ganeshie8
anonymous
  • anonymous
And I call the first master : Please save us : @mukushla ..
freckles
  • freckles
\[p+q=-p \\ pq=q \] these are the equations agreed?
mathmath333
  • mathmath333
yes
freckles
  • 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
  • freckles
I think though you say we have another solution
freckles
  • freckles
so let's see if we can play with the equations more to find it
mathmath333
  • mathmath333
i thhink that is same as of frekles
freckles
  • 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
  • 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
  • mathmath333
but wolfram and my book does say \(p=-1/2\) also.
freckles
  • freckles
can i see the link to wolfram
freckles
  • freckles
because my wolfram checks my solutions just fine
freckles
  • freckles
http://www.wolframalpha.com/input/?i=p%2Bq%3D-p+%2C+pq%3Dq
mathmath333
  • mathmath333
http://www.wolframalpha.com/input/?i=solve+2p%5E2%2Bq%3D0%2Cq%5E2%2Bpq%2Bq%3D0+over+reals
freckles
  • 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
  • anonymous
If you use both p and q values, then also I think both equations will not get satisfied..
mathmath333
  • mathmath333
but vieta's formula shouls work too, something is missing
freckles
  • 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
  • mathmath333
:D
freckles
  • freckles
but I would like to know why it isn't working for veita's formula :(
mathmath333
  • mathmath333
thats a sign of good mathmatician.
freckles
  • 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
  • freckles
so I think p=-1/2 shouldn't be a solution
mathmath333
  • mathmath333
:(
freckles
  • freckles
so veita's formula I think does work and the other way gave too many solutions (solutions that needed to be checked)

Looking for something else?

Not the answer you are looking for? Search for more explanations.