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What value of \(k\) makes the expression \(x^2-8x+k=0\) a perfect square?



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  • \(b^2-4ac=0\)
    \(b=4ac\)
    Where \(b=-8,a=1\) and \(c=k\)
    \(-8^2=4×1×k\)
    \(64=4k\)
    Then, dividing both sides by 4
    \(k=16\)

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  • To make an expression a perfect square, add the square of half the coefficient of x.
    Therefore k = the square of half the coefficient of x.
    \(k = \left( \cfrac{-8}{2} \right)^2 = (-4)^2 = 16\)

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