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Grade 10 Math exercises?

  1. Jan 1, 2005 #1
    Hi
    Does anyone know any good sites that offer Grade ten math exercises? It's for my sister. Canadian grade ten math, she's doing stuff like quadratic equations, factoring, linear equations.
    Also can someone explain to me why when you factor an equation with the quadratic equation the answers signs are reversed from when you do it manually? Like when I do it manually I'll get x= -5, 4 but when I do it with the quadratic equation it becomes x= 5, -4
    Thanks! I don't know how to explain it to my sis!

    Yawie
     
    Last edited by a moderator: Dec 9, 2008
  2. jcsd
  3. Jan 1, 2005 #2
  4. Jan 2, 2005 #3
    Thanks! I'll go check that out!

    About the quadratic thing. I think the right word was quadratic formula...sorry. =P
    for instance:
    lets say for x^2 + 5x + 4
    if i do it by trial and error the answer would be (x+4)(x+1)
    however, if i do it with the quadratic formula...the complicated one. I don't know how to write it... the answer would be (x-4)(x-1) which doesn't work when u multiply them out.
    btw how do u write out those complicated equations?

    Thanks alot!
    Yawie
     
    Last edited by a moderator: Dec 9, 2008
  5. Jan 2, 2005 #4
    edit: oops sorry, misread the question.
     
  6. Dec 9, 2008 #5
    i think i know the answer to your quadratic equation question.
    its because the formula for factored form is y=a(x-s)(x-t), so since its -s and -t, you need to change the sign in front of each x value (or zero) before you put it in the equation manually.
    for example if the quadratic equation gave you the roots x=5 and x=-3, you would need to switch them to -5 and 3 before putting them in the equation like so :
    y=a(x-5)(x+3)
    hope it helped! :D
    - Bradie
     
  7. Dec 9, 2008 #6

    danago

    User Avatar
    Gold Member

    Think about how the factorised form comes about, and what it means to "solve" a quadratic.

    By using the quadratic formula, you are essentially solving the following equation:

    x^2 + 5x + 4 = 0

    Which, as you obtained, has the two solutions x=-4 and x=-1.

    Now when a quadratic equation is in the form:

    (x-a)(x-b) = 0,

    We know that EITHER x-a=0 or x-b=0 (or maybe both), since the product of two numbers can only be zero if one or both numbers are zero. Solving these two equations yields the solutions x=a and x=b. Can you now see why the sign reverses?

    So in the example you gave, solving for its zeros gives the two solutions x=-4 or x=-1, therefore any factored form should have the same two zeros. If you use the one without changing the signs:

    (x-4)(x-1),

    and try plugging either x=-1 or x=-4 into it, you will see that it doesnt equal zero, so it cannot be the same quadratic.
     
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