[Noilemaillilm]

Given the equation x^2+kx+1=0 and x^2-x +k=0 has common roots, find the values of k and the common roots.

Somebody asked me for help on this question but I'm kinda stumped as well. Anyway here's what I came up with, I have a feeling that I'm wrong though.

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x^2 + kx + 1 = 0  b^2 - 4ac = 0  k^2 - 4 (1)(1) k^2 = 4 k = 2  x^2 + 2x + 1 = 0                          (x+1) (x+1) = 0                                                                                                                      x = -1  ----------------------------------------------------  X^2 - x + k = 0  b^2 - 4ac = 0  (-1)^2 - 4(1)(k) 1 - 4k = 0 -4k = -1 k = 1/4  x^2 - x + 1/4 [ -b +/- sqrt b^2 - 4ac ] / 2a [ -(-1) +/- sqrt (-1)^2 - 4(1)(1/4) ] / 2(1) 1 +/- 0 / 2 x = 1/2

Is my answer correct?

[WomanBreast40356]

no idea!

can you substitute k=x^2-x into x^2+kx+1=0 ie x=1 k=0

?

[portoskins]

For a general quadratic equation, ax^2 + bx^2 + c = 0:
roots, x = (-b +/- sqrt(b^2 - 4ac))/2a

So, solving for the two equations, the common root would be
(-k +/- sqrt(k^2-4))/2 = (-1 +/- sqrt(1-4k))/2

It turns out
k = 1 satisfies the equation

And so the common roots x = (1 +/- i*sqrt(3))/2

...I think.

[Noilemaillilm]

Hmm... so there aren't any sure answers to the question then?

[portoskins]

Hmm... so there aren't any sure answers to the question then?

This is your homework not mine :P

I didn't bother to verify my answers, but it's probably simple to do so (just plug the numbers back in) and if I'm right, you have at least a subset of the solution haha

[Noilemaillilm]

This is your homework not mine :P

I didn't bother to verify my answers, but it's probably simple to do so (just plug the numbers back in) and if I'm right, you have at least a subset of the solution haha

I tried plugging back using your answers and they didn't equal to 0. I assume that the solution will be correct when plugging back will give an answer of 0?

I've tried plugging back mine and they equal to 0. Does that mean that my answer is correct? I'm not sure myself .

[portoskins]

Whoops I think the roots are supposed to be x = (1 +/- i*sqrt(3))/2

I had an extra (-) there

edit: BTW, your answers are incorrect in the sense that you did not have common roots that satisfy both equations.

edit: and yet...I found an error in my post above. Sigh. Will try to work it out.

For a general quadratic equation, ax^2 + bx^2 + c = 0:
roots, x = (-b +/- sqrt(b^2 - 4ac))/2a

So, solving for the two equations, the common root would be
(-k +/- sqrt(k^2-4))/2 = (-1 +/- sqrt(1-4k))/2 [ERROR]

It turns out
k = 1 satisfies the equation

And so the common roots x = (1 +/- i*sqrt(3))/2

...I think. Unfortunately k = 1 is *not* the solution. I mostly did it by inspection (and I made and error)...so now I'm hand-calculating this and I've reached a point where I don't feel like working out the algebra. Perhaps I'm misinterpreting the term "common roots".

I don't feel like solving the following:

(-k +/- sqrt(k^2 - 4))/2 = (1 +/- sqrt(1-4k))/2

[Noilemaillilm]

I know how you feel man, this isn't my homework but it kinda bugs me for not being able to solve it. Thanks for your help mate. In the meantime I'll try to ask somebody else and hope that someone here would know the solution to this problem.

[portoskins]

Yeah it's annoying me now.

Feels like it's too complicated, so I looked for some way to verify the existance of common roots. Then I found this online:

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Theorem: Consider the two quadratic equations a m2 + b m + c = 0 a'm2 + b'm + c'= 0 with a and a' not zero, together with 3 determinants             |b     c|          |a     c|         |a     b|    A =  |       |    B = - |       |    C =  |       |         |b'   c'|          |a'   c'|         |a'   b'|     The equations have a common root if and only if A.C = B2. The expression A.C = B2 is called the eliminant of Sylvester.  Proof: We multiply the first equation with a', the second one with a, and we calculate the difference of these two results. (ba' - a'b) m + (a'c - ac') = 0. From this, the system of the two quadratic equations is equivalent with    / (ba' - a'b) m + (a'c - ac') = 0. \ a m2 + b m + c = 0     These equations have a common root if and only if the root of the first one, is a root of the second equation. The root of the first equation is (a'c - a c') / (b'a - a b') = B/C . The condition then is:           a (B/C)2 + b (B/C) + c = 0      a B2 + b B C + c C2 = 0      a B2 = -C ( b B + c C)      ...      a B2 = a ( C A)  and since a is not 0      A.C = B2

If I worked out the math right, the two quadratic equations won't meet this requirement (B2 really means B^2 = B.B but the page didn't paste correctly). Perhaps it's late and I suck at math.

Check with your friend to make sure he gave you the right problem?

edit: see next post (I misinterpreted the theorem above...I thought A, B, and C were matrices when they were actually determinants :/).

[viagbloggerz]

[offtopic] Ive just realised somthing...




















Im thick[thumbdown]

[DevaRextusidis]

Well, I tried doing it in maple. I got one real answer and two complex answers for k. However the real answer is really complex and has stuff like thrid roots in it so maybe its wrong...



First thing i did was work out roots in terms of k. Then i tried putting each root equal to another from the other equation to see if anything came out. "k1" was my real answer, putting that back into the equations they shared an identical root.

so if k = -2.658967080 and x=2.205569430 then both equations are zero.

[portoskins]

Phew, good thing I didn't try working out the math...would have been terrible since I don't have a symbolic evaluation software at home. Thanks Sink!

And now that I reworked the problem with an alternate method (Slyvester's eliminant) I got the same k value

[Noilemaillilm]

I'm going nuts. My friend just got back to me about the answer that I asked her for.

She said the answer is k = -2 and x = -1.

Damn, I give up.[surrender] Anyway, thanks guys for trying to help. Appreciate it a lot.

[Blelidupgerie]

I'm going nuts. My friend just got back to me about the answer that I asked her for.

She said the answer is k = -2 and x = -1.

Damn, I give up.[surrender] Anyway, thanks guys for trying to help. Appreciate it a lot.

That's wrong... it gives x^2 + 3 for the first equation and x^2 - 1 for the second.

.... in your own solution... what makes you think b^2 - 4ac = 0???

it must be greater or equal to 0 because it has to be rooted - nothing says it has to be 0.