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HELP ASAP! Find the zeros of y = x^2 - 8x - 3 by completing the square.

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Notice (x+4)2 = x2 + 8x +16 which differs from the question by a constantSo we can write: x2 + 8x + 3 = (x+4)2 - 13 (x+4)2 -13 = 0(x+4)2 = 13 (by adding 13)x+4 = +-sqrt(13) (square root remembering to include the +-)x = -4 +-sqrt(13) (subtracting 4)So we have answers of:x = - 4 + sqrt(13)x = - 4 - sqrt(13)which can both be checked by substitution into the or
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