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mathmath333
 one year ago
quadratic equation
mathmath333
 one year ago
quadratic equation

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align}& \normalsize \text{If}\ p\ \text{and}\ q\ \text{are the roots of the equation} \hspace{.33em}\\~\\ & x^2+px+q=0,\normalsize \text{then find the value of}\ p.\ \hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align}& p+q=p\hspace{.33em}\\~\\ & pq=p \hspace{.33em}\\~\\ \end{align}}\) from the sum and product of roots i got this , is this correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For equation : \(ax + by + c = 0\), Sum of Roots = \(\frac{b}{a}\) Product of Roots = \(\frac{c}{a}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you sure p = 1??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If p = 1 then from equation 2, q = 1.. But from equation 1, 2p = q does not get satisfied.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This looks good.. :)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1lol 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
 one year ago
Best ResponseYou've already chosen the best response.00 and 1/2 are all right, but how p = 1 is possible?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1how did u find \(0\) as a solution.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, there is a slight mistake that you have done..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I also being mad, I did not check that.. What you got for Product of Roots?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[pq = q\] This must be the product no??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0See, the formulae I gave above, product of roots = c/a.. in your question c = q and a = 1..

Empty
 one year ago
Best ResponseYou've already chosen the best response.0The only way \(pq=p\) can be true is if: \(q=1\) OR \(p=0\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1ok how about \(p=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Empty that should be q on right side, not p..

Empty
 one year ago
Best ResponseYou've already chosen the best response.0? \(pq=p\) is the same thing as \(p=qp\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just solve these equations: \[p + q = p \\ pq = q\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ohh I see, I switched out my p's and q's. Oh well haha, same idea really since they're completely symmetric. XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathmath333 if you use pq = p (the original one you did above), p = 1 as solution is not possible in any case..

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1ok i got \(p=0,1\) how about \(p=1/2\) now ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That should be pq = q...

Empty
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align}& p+q=p\hspace{.33em}\\~\\ & pq=p \hspace{.33em}\\~\\ \end{align}}\) What mathmath wrote is wrong here, I went off this, but really the last equation is what @waterineyes is trying to tell us!

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1why \(p=1\) is not possible ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0p=1 is possible only when q=0. So you have to work out the consequence of that in the other equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hey @mathmath333 listen to me, what is product of roots you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now, you get the cases of 0 and 1 ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, we left with how p = /12 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Put p = 1/2 in first equation and find q...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But it is not satisfying the other equation.. :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What 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
 one year ago
Best ResponseYou've already chosen the best response.0\[pq = q \\ \frac{1}{2} \times 1 = 1 \\ \frac{1}{2} \ne 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Empty if you have something to correct or say, then you can proceed, I have tried my bit. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah waiting, ordering coffee or tea something? :P

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1is \(0\) and \(1\) both correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@SolomonZelman your help is needed here..

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1he came first but left

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes 0 and 1 both are satisfying but 1/2 is not..

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1seee this wolfram link http://www.wolframalpha.com/input/?i=solve+2p%5E2%2Bq%3D0%2Cq%5E2%2Bpq%2Bq%3D0+over+reals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now wolfram will tell the secrets of us that we don't know anything. :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I call the first master : Please save us : @mukushla ..

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[p+q=p \\ pq=q \] these are the equations agreed?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ the second equation we have } \\ pq=q \\ pqq=0 \\ q(p1) =0 \implies q=0 \text{ or } p=1 \\ \text{ going back to first equation } \\ \text{ if } q=0 \text{ we have } p=p \text{ which forces } p=0 \\ \text{ if } p=1 \text{ then } 1+q=1 \implies q=2 \\ \text{ so far we have the following pairs of solutions } \\ (p,q) \\ (0,0) \\ (1,2)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I think though you say we have another solution

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so let's see if we can play with the equations more to find it

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i thhink that is same as of frekles

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[p=\frac{1}{2} \text{ trying first } \\ \frac{1}{2}+q=(\frac{1}{2}) \\ \frac{1}{2}+q=\frac{1}{2} \\ q=1 \\\text{ so if } p=\frac{1}{2} \text{ is a solution } \\ \text{ then we should get } q=1 \text{ for the second equation too} \\ \frac{1}{2}q=q \\ \frac{1}{2}qq=0 \\ q=0 \\ \text{ so } p=\frac{1}{2} \text{ doesn't seem to work out }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the only solutions I think there to be is (0,0) and (1,2) where those ordered pairs are in the form (p,q)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but wolfram and my book does say \(p=1/2\) also.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2can i see the link to wolfram

freckles
 one year ago
Best ResponseYou've already chosen the best response.2because my wolfram checks my solutions just fine

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=solve+2p%5E2%2Bq%3D0%2Cq%5E2%2Bpq%2Bq%3D0+over+reals

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ok 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
 one year ago
Best ResponseYou've already chosen the best response.0If you use both p and q values, then also I think both equations will not get satisfied..

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but vieta's formula shouls work too, something is missing

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[2p^2+q=0 \\ q^2+pq+q=0 \\ 2p^2=q \\ 2p^2=q \\ (2p^2)^2+p(2p^2)+(2p^2)=0 \\ 4p^42p^32p^2=0 \\ 2p^2(2p^2p1)=0 \\ p=0 \text{ or } 2p^2p1=0 \\ p=0 \text{ or } (2p+1)(p1)=0 \\ p=0 \text{ or } p=\frac{1}{2} \text{ or } p=1 \] hmm... I don't know why vieta's formula isn't working for that one p

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but I would like to know why it isn't working for veita's formula :(

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1thats a sign of good mathmatician.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=x%5E21%2F2x1%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
 one year ago
Best ResponseYou've already chosen the best response.2so I think p=1/2 shouldn't be a solution

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so 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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