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    I've been given this question by a friend and I'm stuck on it.

    The roots of the equationx^2+(p-10)x=10pare a and a+4. Find the possible values of p.

    I started by expanding the bracket but I have no idea if that's right. I can sort of see what it's getting at though.

    What confused me is the use of the letter a, since the formula for quadratics is ax^2..., I thought a=1 but I don't think it would be that easy...

    Edit: This is what I have now:

    x^2+px-10x-10p=0
    x(x-10)+p(x-10)=0
    Not sure what to do from here

    Would I use x=10?
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    x^2 + (p-10)x - 10p = 0 is how it would look in your more 'general' form I presume you are use to.
    here the roots are x = a and x = a+4

    Put these in to the equation one at a time and solve for p (remember you are trying to find p not a)
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    (Original post by Bude8)
    I've been given this question by a friend and I'm stuck on it.

    The roots of the equationx^2+(p-10)x=10pare a and a+4. Find the possible values of p.

    I started by expanding the bracket but I have no idea if that's right. I can sort of see what it's getting at though.

    What confused me is the use of the letter a, since the formula for quadratics is ax^2..., I thought a=1 but I don't think it would be that easy...

    Edit: This is what I have now:

    x^2+px-10x-10p=0
    x(x-10)+p(x-10)=0
    Not sure what to do from here

    Would I use x=10?
    Taking out the common factor gives you (x+p)(x-10)
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    (Original post by TenOfThem)
    Taking out the common factor gives you (x+p)(x-10)
    Wait, how did you get that?
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    (Original post by Bude8)
    I've been given this question by a friend and I'm stuck on it.

    The roots of the equationx^2+(p-10)x=10pare a and a+4. Find the possible values of p.

    I started by expanding the bracket but I have no idea if that's right. I can sort of see what it's getting at though.

    What confused me is the use of the letter a, since the formula for quadratics is ax^2..., I thought a=1 but I don't think it would be that easy...

    Edit: This is what I have now:

    x^2+px-10x-10p=0
    x(x-10)+p(x-10)=0
    Not sure what to do from here

    Would I use x=10?
    Use the Viéte formula
    For ax^2+bx+c=0 where the two roots are x_1, x_2, then
    x_1+x_2=\frac{-b}{a}
    x_1\cdot x_2=\frac{c}{a}
    From above for example
    (x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=\left (-\frac{b}{a}\right )^2-\frac{4c}{a}
    For your question
    a=1 b=p-10 c=-10p and (x_1-x_2)^2=16
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    (Original post by Bude8)
    Wait, how did you get that?
    You had

    x(x-10) + p(x-10)

    Common factor
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    (Original post by TenOfThem)
    You had

    x(x-10) + p(x-10)

    Common factor
    So the (x-10) cancel out, but aren't you left with just x+p?
    I don't get why there's still a (x-10)
    I probably need to revise factorising again...
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    (Original post by Bude8)
    So the (x-10) cancel out, but aren't you left with just x+p?
    I don't get why there's still a (x-10)
    I probably need to revise factorising again...
    I do not understand how you think they "cancel out"

    It is like

    ax+bx = x(a+b)

    The x doesn't "cancel"
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    (Original post by TenOfThem)
    I do not understand how you think they "cancel out"

    It is like

    ax+bx = x(a+b)

    The x doesn't "cancel"
    ok so from x(x-10)+p(x-10)=0
    you get
    (x+p)(x-10)

    What I don't understand is why one of the (x-10) goes...
    So confused right now
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    (Original post by Bude8)
    ok so from x(x-10)+p(x-10)=0
    you get
    (x+p)(x-10)

    What I don't understand is why one of the (x-10) goes...
    So confused right now
    x(x-10)+p(x-10)=(x+p)(x-10) since if you take out a common factor of (x-10) you'll notice that you times this by x and p which the same as (x+p)(x-10), basically if you look there you get p(x-10) and x(x-10), look at the bracket and see why it makes sense

    It's not gone, you've just factored it by using foil or otherwise you can see that (x+p)(x-10) gives the exact same result
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    (Original post by Robbie242)
    x(x-10)+p(x-10)=(x+p)(x-10) since if you take out a common factor of (x-10) you'll notice that you times this by x and p which the same as (x+p)(x-10), basically if you look there you get p(x-10) and x(x-10), look at the bracket and see why it makes sense
    Ah, I see, thank you!

    Rep for everyone who has helped, I've solved it now.
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    (Original post by Bude8)
    ok so from x(x-10)+p(x-10)=0
    you get
    (x+p)(x-10)

    What I don't understand is why one of the (x-10) goes...
    So confused right now
    Then I agree

    You need to revise factorising/multiplication of brackets


    In your first post you went from (p-10)x to px - 10x
    Did you wonder where the "extra" x came from
    This is the same question!
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    (Original post by Bude8)
    Wait, how did you get that?
    the distributivity

    (x-10)[A+B]=A(x-10)+B(x-10)

    in reverse order is a factorizing
    A(x-10)+B(x-10)=(x-10)(A+B)
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    (Original post by Bude8)
    Ah, I see, thank you!

    Rep for everyone who has helped, I've solved it now.
    good
 
 
 
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