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Math Help - Function

  1. #1
    Junior Member
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    Post Function

    Hi there, I have a problem that I can not seem to figure out. It is asking on the condition that a >0 to prove that the quadratic expression ax^2 +bx +c is positve for all real values of x when b^2 < 4ac. then it saids based on that to find the range of values of p for which the quadratic function of x

     <br />
f(x) \equiv 4x^2 + 4px - (3p^2 +4p -3)<br />

    is positive for all real values of  x illustrate your result by making sketch graphs of f(x) for each of the cases p=0 and p=1

    Can some one out there help me to figure out this problem/
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  2. #2
    Super Member Bacterius's Avatar
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    First question : First, you know that a> 0. That means that if you sketch the graph for any permitted value of p, it will be a curved parabola open to the top. Therefore, there are only two possible situations : either it never touches the x-axis, therefore there are no real solutions and the equation is always positive, either it cuts the x-axis and there are solutions. But the equation cannot be all negative. Now, if b^2 < 4ac, that means that b^2 - 4ac < 0. Basically, it means that the discriminant is negative, thus the equation admits no real solution and therefore is positive for all real x (you got to rewrite it properly though, I gave you the idea).

    Second question : you need to find all values of p for which the given quadratic has only positive values (that is, has no real solution). Think of your equation like this :

    ax^2 + bx + c

    Where :

    a = 4
    b = 4p
    c = -(3p^2 + 4p - 3)

    Get the discriminant of this equation :

    \Delta = b^2 - 4ac

    By substituting your values :

    \Delta = (4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)]

    Now, what is the characteristic feature of a quadratic equation with no real solution (i.e. all values are positive, in our case) ? The discriminant is negative. Therefore you have :

    \Delta < 0

    (4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)] < 0

    16p^2 + 16 \times (3p^2 + 4p - 3) < 0

    16p^2 + 48p^2 + 64p - 48 < 0

    64p^2 + 64p - 48 < 0

    Solve for p and you will then have the range of p when the quadratic equation has no real solution (always positive).

    Does it help ? Ask if you still need help.

    PS : why did you put a congruent sign ( \equiv) on your function ?
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  3. #3
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    Quote Originally Posted by Bacterius View Post
    First question : First, you know that a> 0. That means that if you sketch the graph for any permitted value of p, it will be a curved parabola open to the top. Therefore, there are only two possible situations : either it never touches the x-axis, therefore there are no real solutions and the equation is always positive, either it cuts the x-axis and there are solutions. But the equation cannot be all negative. Now, if b^2 < 4ac, that means that b^2 - 4ac < 0. Basically, it means that the discriminant is negative, thus the equation admits no real solution and therefore is positive for all real x (you got to rewrite it properly though, I gave you the idea).

    Second question : you need to find all values of p for which the given quadratic has only positive values (that is, has no real solution). Think of your equation like this :

    ax^2 + bx + c

    Where :

    a = 4
    b = 4p
    c = -(3p^2 + 4p - 3)

    Get the discriminant of this equation :

    \Delta = b^2 - 4ac

    By substituting your values :

    \Delta = (4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)]

    Now, what is the characteristic feature of a quadratic equation with no real solution (i.e. all values are positive, in our case) ? The discriminant is negative. Therefore you have :

    \Delta < 0

    (4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)] < 0

    16p^2 + 16 \times (3p^2 + 4p - 3) < 0

    16p^2 + 48p^2 + 64p - 48 < 0

    64p^2 + 64p - 48 < 0

    Solve for p and you will then have the range of p when the quadratic equation has no real solution (always positive).

    Does it help ? Ask if you still need help.

    PS : why did you put a congruent sign ( \equiv) on your function ?
    I think that is how they have it in the book, I am guesing that is what it is.
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  4. #4
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    Post

    Quote Originally Posted by scrible View Post
    I think that is how they have it in the book, I am guesing that is what it is.
    Thank you very much. I really appreciate it
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  5. #5
    Super Member Bacterius's Avatar
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    Quote Originally Posted by scrible View Post
    Thank you very much. I really appreciate it
    No problem, have a nice day !
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