The ages of three family children can be expressed as consecutive integers. the square of the age of the youngest child is 4 more than eight

Question

The ages of three family children can be expressed as consecutive integers. the square of the age of the youngest child is 4 more than eight times the age of the oldest child. find the ages of the three children.

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    2021-10-14T01:54:52+00:00

    Let the middle child be x
    youngest child = x – 1
    Oldest child = x + 1

    (x – 1)^2 = 8(x + 1) + 4
    x^2 – 2x + 1 = 8x + 8 + 4 
    x^2 – 10x – 11 = 0
    (x – 11) * (x + 1) = 0

    X = – 1 makes no sense. How can a middle child be – 1 years old?
    x = 11

    The youngest child is 10
    The middle child is 11
    The oldest child is 12

    Check
    =====
    10^2 = 100
    (8*12) + 4
    96 + 4 = 100 So the results have been checked.

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