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    madmaths's Avatar
    madmaths Posts: 6, Reputation: 1
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    #1

    Mar 7, 2011, 08:03 AM
    Expand the brackets in the following expression
    (5y-3)^2

    ------------------
    Factorise the following expressions

    x^2 - x - 12 (2)


    9y^2-25

    I've several questions like this if someone could let me know what to do with this one then maybe I could get an idea of what to do with the rest. Thanks!

    ------------------
    solve the equation.

    9/(x=7)= 4/(x-3)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Mar 7, 2011, 08:44 AM

    1. For this one, remember that:



    2. (a) Let's say the factorised expression is (x - a)(x - b), when you expand it out, you get:



    Now, (a + b) is 1 from the given and ab = -12

    can you guess what a and b can be?

    (b) Use the difference of two squares, that is, if you have:



    3. I guess you mean:



    ?

    Cross multiply, then move all the terms in x and the terms without x on separate sides of the equation. Last, divide by the coefficient of x.

    Post what you get! :)
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    madmaths Posts: 6, Reputation: 1
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    #3

    Mar 8, 2011, 04:00 AM
    thanks going by that the answer I work out is x=-1/5 .
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Mar 8, 2011, 04:08 AM

    For the last question?

    Hm... I'm getting something else.

    Cross-multiplying:



    Expanding:



    Move around:







    Is that okay now? :)
    madmaths's Avatar
    madmaths Posts: 6, Reputation: 1
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    #5

    Mar 8, 2011, 04:20 AM
    Hi yes sorry it was the last question.

    I cross multiplied:
    9*(x-3)=4*(x-7) I notice you kept the + sign in thesecond bracket, I thought it became negative.
    so then I worked out
    9x-27=4x-28
    Then I collected the variable x on the left side and the rest on the right
    9x-4x=27-28
    5x=-1
    then I divided the right side with the coefficient of x
    x=-1/5
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    madmaths Posts: 6, Reputation: 1
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    #6

    Mar 8, 2011, 04:24 AM
    For question 1.
    I get
    (5y-3)^2=(5y-3)(5y-3)
    (5y-3)(5y-3)=(5y)^2-(2.3.5y)+3^2
    (5y)^2-(2.3.5y)+3^2=25y^2-30y+9
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    Unknown008 Posts: 8,076, Reputation: 723
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    #7

    Mar 8, 2011, 04:38 AM

    Oh, instead of '=', you meant '-', okay, I thought it was '+' since it's the sign usually replaced by '='.

    Okay, last one is good :)

    First one is good too :)
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    madmaths Posts: 6, Reputation: 1
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    #8

    Mar 8, 2011, 05:03 AM
    Thanks for all your help, I've worked out the other two, hopefully they are right!

    I've worked out 9y^2-25 to be (3x-5((3x=5)

    and x^2-x-12 to be (x-)(x=3)

    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #9

    Mar 8, 2011, 05:14 AM

    Oh well, that's not too easy to understand.

    9y^2 - 25 = (3y + 5)(3y - 5)

    x^2 - x - 12 = (x - 4)(x + 3)

    :)
    RPVega's Avatar
    RPVega Posts: 29, Reputation: 2
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    #10

    Mar 9, 2011, 11:56 AM
    (5y-3)^2 = (5y-3)(5y-3)=25y - 15y -15y + 9 = 25y - 30y + 9
    Remember the word "FOIL": First, Outer, Inner, Last
    Multiply the First terms of the two binomials: 5y * 5y = 25y
    Multiply the Outer terms of the two binomials: 5y * -3 = -15y
    Multiply the Inner terms of the two binomials: -3 * 5y = -15y
    Multiply the Last terms of the two binomials: -3 * -3 = 9
    Finally, add all the terms together: 25y + (-15y) + (-15y) + 9 = 25y -30y + 9

    9/(x-7) = 4/(x-3)
    4(x-7) = 9(x-3)
    4x - 28 = 9x - 27
    -28 + 27 = 9x - 4x
    -1 = 5x
    x = -1/5
    RPVega's Avatar
    RPVega Posts: 29, Reputation: 2
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    #11

    Mar 9, 2011, 02:05 PM
    Sorry, I made a boo,boo. (5y-3)^2 = 25y^2-30y+9
    (5y-3)^2 = (5y-3)(5y-3)
    Remember the word "FOIL": First, Outer, Inner, Last
    Multiply the First terms in the two binomials: 5y*5y=25y^2
    Multiply the Outer terms in the two binomials: 5y*(-3)=-15y
    Multiply the Inner terms in the two binomials: (-3)*5y=-15y
    Multiply the Last terms in the two binomials: (-3)*(-3)=9
    Next, add the results of all the above together: 25y^2+(-15y)+(-15y)+9=
    25y^2-30y+9

    Here's a shortcut to the above:

    When taking the square of a binomial, (a+b)^2, take the square of the first
    term, a^2, plus the product of the two terms multiplied by 2, 2ab, plus
    the square of the last term, b^2. The result is: (a+b)^2=a^2 + 2ab + b^2.
    Try this with the above example: (5y-3)^2. With this shortcut, you can arrive
    at the answer in ONE step: (a+b)^2=(5y+(-3))^2=(5y)^2-(5y*(-3)*2)+(-3)^2=
    25y^2-30y+9, where a=5y; and b=(-3). Good luck! The U.S. needs more engineers and math teachers!
    RPVega's Avatar
    RPVega Posts: 29, Reputation: 2
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    #12

    Mar 9, 2011, 02:16 PM
    9y^2-25=(3y+5)(3y-5)
    Here's a shortcut:
    (ay+b)(ay-b)=(ay)^2-b^2. This applies when multiplying two binomials where the b terms
    are of opposite signs in the two binomials. In the above example, a=3; and b=5.
    Good luck!

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