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Algebra II

  1. Mar 29, 2006 #1
    I'm supposed to use this equation:

    y = x^2-8x+7

    To solve the following questions:

    1) What is the value of the discriminant?
    2) Find the roots by factoring and solving.
    3) Find the roots by using the quadratic equation.

    Can anyone give me some help?

    Thanks! :)
     
  2. jcsd
  3. Mar 29, 2006 #2

    JasonRox

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    Try searching on google.

    Do you know what the discriminant is?

    Do you know how to factor?

    Do you know what the quadratic equation is?

    I can help you, but any type of help would be giving the answer.

    http://home.alltel.net/okrebs/page6.html

    Here's a link if you don't feel like searching.
     
  4. Mar 29, 2006 #3

    JasonRox

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    Is this really Algebra II?
     
  5. Mar 29, 2006 #4
    Tryed myself, the discriminant is "x^2 - 8x"?

    The factor is: (x - 7)(x - 1) But are those the roots also?

    Not sure about #3.

    Thanks for the help! :)

    PS: Yes, this is in my Algebra II class.
     
  6. Mar 29, 2006 #5

    JasonRox

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  7. Mar 29, 2006 #6

    HallsofIvy

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    Do you know the quadratic formula? I'll bet it's in your book. I'll also bet they define "discriminant" in the same section. (No, it's not "x^2- 8x". The discriminant is a number not an expression in x.)
     
  8. Mar 29, 2006 #7

    JasonRox

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    Why look in the book when people online will answer it? :uhh:
     
  9. Mar 29, 2006 #8
    It probably is in my book, but I'm sick and my book was left in my locker as I thought I was going to school today. I'm looking it up on how to do it, although I can find examples I can't figure out how they change it.

    Thanks :)

    PS: The roots are x = {7,1} I believe?
     
  10. Mar 29, 2006 #9
    The quadratic formula is:

    {-8 ± Sq Root (64 - 28)} ÷ 2

    That means the discriminant is:

    64 - 28

    I hope thats right.

    Thanks! :blushing:
     
  11. Mar 29, 2006 #10
    Thanks for all the help. :)

    On the question:

    Find the roots by using the quadratic equation.

    Would the answer be x = {7,14}?

    Thanks :)
     
  12. Mar 29, 2006 #11
    Substitute your answers in the original equation. You'll see immediately if they're correct.
     
  13. Mar 29, 2006 #12
    Okay, thanks. :)

    How would I find the 'x =' in the equation? (y = x^2-8x+7)

    THANKS! :)
     
  14. Mar 29, 2006 #13

    Tom Mattson

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    High school and college level homework goes into the Science Education Zone, please.

    You have to show some work first.
     
  15. Mar 29, 2006 #14

    JasonRox

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    Hint: Take a closer look at the Quadratic Formula.
     
  16. Mar 30, 2006 #15

    HallsofIvy

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    The quadratic formula is "overkill" for this one. How can you factor 7?

    Edit: I just noticed, this is not x^2- 8x+ 7= 0, which would give x= 7, x= 1 as roots, but y= x^2 - 8x+ 7. To solve that for x, in terms of y, you would need to use the quadratic formula (or complete the square).
     
    Last edited: Mar 31, 2006
  17. Mar 30, 2006 #16
    discriminant = #'s inside the radical.

    to solve the factoring problem you must know :
    F-irst
    O-utside
    I-nside
    L-ast

    For example if we had [tex]x^2+7x+12=0[/tex] we would factor it light this
    [tex](x+3)(x+4)[/tex]

    To know if you have done your work right use FOIL
    First: [tex]x \times x[/tex]
    Outside: [tex]4 \times x[/tex]
    Inside: [tex]3 \times x[/tex]
    Last: [tex]3 \times 4[/tex]


    To find the zeros (or answers) of x for : [tex](x+3)(x+4)[/tex] we would simply take:
    [tex](x+3)=0[/tex] and solve for x and [tex](x+4)=0[/tex]
     
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