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28 July, 20:35

How to solve Y^2-8y-16

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  1. 28 July, 22:06
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    You can't solve it because it isn't equal anything

    but we can factor

    since the quadratic coefient (number in front of y^2 term) is 1

    we can do this

    for ax^2+bx+c

    when a=1

    what 2 numbers multiply to c and add to b

    what 2 numbers multiply to - 16 and add to - 8

    no numbers

    if you had y^2-8y+16 then that would be (x-4) (x-4)

    but no

    ok, so we need to complete the squaer

    so

    y^2-8y-16

    take 1/2 of linear coefient and square it

    -8/2=-4, - 4^1=16

    add negative and poisitve to it

    y^2-8y+16-16-16

    factor

    (y-4) ^2-16-16

    (y-4) ^2-32

    then we can force factor again

    remember difference of 2 perfect square

    a^2-b^2 = (a+b) (a-b)

    √32=4√2

    so

    (y-4) ^2 - (4√2) ^2 = (y-4+4√2) (y-4-4√2)

    the factore form would be (y-4+4√2) (y-4-4√2)

    not equal to anything tho so we can't solve
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