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Math Help - Need Help with problems

  1. #1
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    Need Help with problems

    I missed class and I am totally lost with these problems. What we have to do is find all real zeros of the function algebraically. My teacher had this equation up on the board but I can;t find it anywhere so any help will help
    -bsqaurerooe+-4acdivided by 2a

    f(x)=x^2-16

    f(x)=t^2-4t^2+4t

    g(t)=t^5-6t^3+9t
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  2. #2
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    Quote Originally Posted by jvrivera View Post
    I missed class and I am totally lost with these problems. What we have to do is find all real zeros of the function algebraically. My teacher had this equation up on the board but I can;t find it anywhere so any help will help
    -bsqaurerooe+-4acdivided by 2a

    f(x)=x^2-16

    f(x)=t^2-4t^2+4t

    g(t)=t^5-6t^3+9t
    x^2 - 16 = 0

    x^2 = 16

    x = \pm 4.


    t^2 - 4t^2 + 4t = 0

    4t - 3t^2 = 0

    t(4 - 3t) = 0

    t = 0 or 4 - 3t = 0

    t = 0 or t = \frac{4}{3}.


    t^5-6t^3+9t = 0

    t(t^4 - 6t^2 + 9) = 0

    t(t^2 - 3)^2 = 0

    t = 0 or t^2 - 3 = 0

    t = 0 or t = \pm \sqrt{3}.
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  3. #3
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    Quote Originally Posted by jvrivera View Post
    I missed class and I am totally lost with these problems. What we have to do is find all real zeros of the function algebraically. My teacher had this equation up on the board but I can;t find it anywhere so any help will help
    -bsqaurerooe+-4acdivided by 2a
    I think you mean the "quadratic formula" which is
    "if ax^2+ bx+ c= 0, then x= \frac{-b\pm \sqrt{b^2- 4ac}}{2a}" (you forgot the b^2).

    f(x)=x^2-16
    As Prove It said, you can factor. If you want to use the quadratic formula, a= 1, b= 0, c= -16. It is always better to factor if you can. Save the quadratic formula for problems where you can't factor.

    f(x)=t^2-4t^2+4t
    This is probably NOT what you intend. This is -3t^2+ 4t= t(-3t+ 4)= 0 which has roots t= 0 and t= 4/3. But I suspect you meant t^2- 4t+ 4= (t- 2)^2= 0 which has the root t= 2.
    Again, you could use the quadratic formula with a= 1, b= -4, c= 4.

    g(t)=t^5-6t^3+9t
    You can factor this by taking out a single t, t(t^4- 6t^2+ 9)= 0 which has t= 0 as a root. Now let x= t^2 and the other factor becomes x^2- 6x+ 9= (x- 3)^2= 0. That has the single root x= 3 so t^2= 3 and then t= \pm\sqrt{3}
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