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28 April, 17:02

Which values for b make the polynomial 16z2 + bz + 81 a perfect square trinomial?

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  1. 28 April, 20:44
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    As (a+b) ^2 = a^2 + 2*a*b + b^2 ... (i)

    here we have

    16z^2 + bz + 81 which can be written as (4z) ^2 + bz + 9^2

    from (i) we can see that in order to be perfect square bz should be

    bz = 2 * (4z) * (9)

    bz = 72z

    dividing z on both sides we get

    b=72
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