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    Just a quick question regarding this example in my book.

    When it comes completing the square and taking out the coefficient of 2 in this example.

    2x^2-10x
    = 2(x^2-5x)
    = 2[(x-5/2)^2 -(5/2)^2]
    = 2[(x-5/2)^2 - 25/4]
    = 2(x-5/2)^2 - 25/2

    How come the answer is 2(x-5/2)^2 - 25/2 and not 2(x-5/2)^2 - 25/4 ?

    Thanks for the help in advance!


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    (Original post by dthomas86)
    Just a quick question regarding this example in my book.

    When it comes completing the square and taking out the coefficient of 2 in this example.

    2x^2-10x
    = 2(x^2-5x)
    = 2[(x-5/2)^2 -(5/2)^2]
    = 2[(x-5/2)^2 - 25/4]
    = 2(x-5/2)^2 - 25/2

    How come the answer is 2(x-5/2)^2 - 25/2 and not 2(x-5/2)^2 - 25/4 ?

    Thanks for the help in advance!


    Posted from TSR Mobile
    Becuase:
    2 \left(\dfrac{5}{2} \right)^2=2 \left(\dfrac{25}{4} \right)=\dfrac{25}{2}
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    It is like that:

     2x^{2}-10x=2(x^{2}-5x)=2(x^{2}-5x+\frac{25}{4}-\frac{25}{4})=2(x-\frac{5}{2})^{2}-\frac{25}{2}

    because of the fact that  2(x^{2}-5x+\frac{25}{4}-\frac{25}{4})=2(x-\frac{5}{2})^{2}-2\frac{25}{4}
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    Cheers guys. Just to double everything outside of the brackets gets multiplied by everything inside the parentheses?


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    Is this correct? ^


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    (Original post by dthomas86)
    Cheers guys. Just to double everything outside of the brackets gets multiplied by everything inside the parentheses?


    Posted from TSR Mobile

    (Original post by dthomas86)
    Is this correct? ^


    Posted from TSR Mobile
    I'm having trouble understanding what you're asking!

    If you mean does a(b + c) = ab + ac, then yes
 
 
 
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