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Math Help - Help with finding a positive and negative root

  1. #1
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    Help with finding a positive and negative root

    Hi all..

    I've a question which requires to me to find a positive and negative real root for values of p...the equation is...

    x^2 +px +8=p

    Find the values of p has one positive and one negative root.

    Many thanks!
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  2. #2
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    Quote Originally Posted by silvercoo View Post
    Hi all..

    I've a question which requires to me to find a positive and negative real root for values of p...the equation is...

    x^2 +px +8=p

    Find the values of p has one positive and one negative root.

    Many thanks!
    1. You have to solve the equation for x using the quadratic formula:

    x^2+px+(8-p)=0~\implies~x=-\frac p2 \pm \frac12 \sqrt{p^2+4p-32}

    2. Since p^2+4p-32 = (p-4)(p+8) you know that p \geq 4

    3. Since the equation should have positive solutions too you know that

    -\frac p2 + \frac12 \sqrt{p^2+4p-32} > 0~\implies~ \frac12 \sqrt{p^2+4p-32} > \frac p2

    4. Square both sides and you'll get p > 8
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  3. #3
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    Hi!

    Many thanks for your help..

    just wondering, why cant I use b^2 - 4ac >0 ?
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  4. #4
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    Quote Originally Posted by silvercoo View Post
    Hi!

    Many thanks for your help..

    just wondering, why cant I use b^2 - 4ac >0 ?
    That's necessary to get real solutions. But:

    Since p \geq 4 you know that -\frac p2 is a negative number. To get a positive solution the value of the square-root must be greater than that. I used exactly this property to get the inequality.
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  5. #5
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    Correct, then i would need to equate it to >0

    so i should obtain (p+8)(p-4)>0

    which would lead to
    p<-8 or p>4...

    Edit
    Pardon me, it should be p>= 4

    Edit, again
    Nope...its still <-8 or p>4...

    sorry about this..
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  6. #6
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    Quote Originally Posted by silvercoo View Post
    Correct, then i would need to equate it to >0

    so i should obtain (p+8)(p-4)>0

    which would lead to
    p<-8 or p>4...

    Edit
    Pardon me, it should be p>= 4

    Edit, again
    Nope...its still <-8 or p>4...

    sorry about this..
    You are correct: There are also positive solutions if p \leq -8.

    Sorry for the confusion.
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  7. #7
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    Quote Originally Posted by silvercoo View Post
    ...
    Edit, again
    Nope...its still <-8 or p>4...

    sorry about this..
    Try p = 5
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  8. #8
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    Hey thanks. I think i managed to figure it out already..

    thanks once again! =)
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  9. #9
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    The product of the roots of the quadratic equation ax˛+bx+c=0 is c/a

    Therefore to get a positive and a negative root for x^2+px+(8-p)=0 you must have a negative product : 8-p < 0 or p > 8
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