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Values in Quadratic Equation with 2 different roots

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the intervals of all possible value of p which the equation equation: (p-1)x^2+4x+(p-4)=0 has two different roots.



    2. Relevant equations

    ax^2+bx+c>0 ??

    3. The attempt at a solution

    (p-1)x^2+4x+(p-4)>0 ??

    How would I go about solving this?
    Is two roots >0 or =0??

    Could I have an example so that I could then solve the problem on my own or steps I need to take in order to solve for the values of p.
     
  2. jcsd
  3. Nov 27, 2011 #2

    ehild

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    Can you solve a quadratic equation? What formula do you use for the solution? In case of p=2 you have the quadratic equation x^2+4x-2. What are the roots?

    ehild
     
  4. Nov 27, 2011 #3
    yes but I dont know with (p-1) and (p+4)

    -b+/- Squareroot b^2 - 4ac over 2a ?

    Do i plug it in that?
     
  5. Nov 27, 2011 #4

    Mark44

    Staff: Mentor

    Yes. Use the quadratic formula with a = p - 1, b = 4, and c = p - 4.
     
  6. Nov 27, 2011 #5
    ohh okay. I used a pgraphing calculator and think I got the answer.
    p< or equal to 4

    and
    p>1
     
  7. Nov 27, 2011 #6

    ehild

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    Are you sure? What are the roots if p=0.5? When p=4.5?

    ehild
     
  8. Nov 27, 2011 #7
    Ohh. I see. I believe my answer now is p<5 and p>0.
     
  9. Nov 27, 2011 #8

    ehild

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    It is better. But why do you think so?
    When has a quadratic equation two different roots?

    ehild
     
  10. Nov 27, 2011 #9
    When it intersects the x axis. It intersects at two different points. So I just plugged in values for p until I began to see two different roots and then narrowed down my answer.
     
  11. Nov 28, 2011 #10

    ehild

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    Yu can not get the exact solution from a graph. By the way, people could (and can) solve quadratic equations without graphing calculators.
    Use the formula that you copied already for the solution of the quadratic equation

    ax2+bx+c=0

    [tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

    Because of the ± in front of the square root, the quadratic equation has two real roots if the expression under the square root (the discriminant) is greater than zero. There is a single solution if it is equal to zero, and no real solution if the discriminant is negative.
    So you need to find those p values which make the discriminant greater than zero:
    [tex]b^2-4ac>0[/tex]
    Plug in the expressions for a, b, and c, and solve the inequality for p.

    ehild
     
  12. Nov 28, 2011 #11
    Thank you so much for your help!!!! i could not have solved it without your help. I get it. Now lets just see how do on my test this Wednesday.
     
  13. Nov 28, 2011 #12

    ehild

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    Good luck!


    ehild
     
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