## mathmath333 one year ago quadratic equation

1. 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}}

2. 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.

3. anonymous

yes surely...

4. anonymous

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

5. anonymous

Are you sure p = 1??

6. mathmath333

$$p=-1/2$$

7. anonymous

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

8. anonymous

This looks good.. :)

9. 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}}

10. anonymous

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

11. mathmath333

how did u find $$0$$ as a solution.

12. anonymous

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

13. anonymous

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

14. anonymous

$pq = q$ This must be the product no??

15. mathmath333

?

16. anonymous

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

17. Empty

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

18. mathmath333

ok how about $$p=1$$

19. anonymous

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

20. mathmath333

oh ur right .

21. Empty

? $$pq=p$$ is the same thing as $$p=qp$$

22. anonymous

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

23. anonymous

How they are same?

24. 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

25. anonymous

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

26. mathmath333

ok i got $$p=0,1$$ how about $$p=-1/2$$ now ?

27. anonymous

That should be pq = q...

28. 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!

29. mathmath333

lol big drama

30. mathmath333

why $$p=1$$ is not possible ?

31. Empty

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

32. mathmath333

its so confusin

33. anonymous

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

34. mathmath333

$$pq=q$$

35. anonymous

yea that is great..

36. anonymous

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

37. mathmath333

yes i do

38. anonymous

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

39. mathmath333

yes

40. anonymous

*-1/2..

41. anonymous

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

42. mathmath333

$$q=1$$

43. anonymous

Good...

44. anonymous

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

45. mathmath333

how ?

46. 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..

47. anonymous

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

48. anonymous

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

49. mathmath333

wait

50. anonymous

Yeah waiting, ordering coffee or tea something? :P

51. mathmath333

is $$0$$ and $$1$$ both correct

52. anonymous

@SolomonZelman your help is needed here..

53. mathmath333

he came first but left

54. anonymous

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

55. mathmath333
56. mathmath333

@waterineyes

57. anonymous

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

58. mathmath333

59. mathmath333

@ganeshie8

60. anonymous

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

61. freckles

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

62. mathmath333

yes

63. freckles

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

64. freckles

I think though you say we have another solution

65. freckles

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

66. mathmath333

i thhink that is same as of frekles

67. freckles

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

68. 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)

69. mathmath333

but wolfram and my book does say $$p=-1/2$$ also.

70. freckles

can i see the link to wolfram

71. freckles

because my wolfram checks my solutions just fine

72. freckles
73. mathmath333
74. 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 ..

75. anonymous

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

76. mathmath333

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

77. 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

78. mathmath333

:D

79. freckles

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

80. mathmath333

thats a sign of good mathmatician.

81. 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

82. freckles

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

83. mathmath333

:(

84. freckles

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