Two positive numbers differ by 5 and the sum of their squares is 193 . find the two numbers

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Two positive numbers differ by 5 and the sum of their squares is 193 . find the two numbers

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Madelyn 3 weeks 2021-10-04T15:38:02+00:00 1 Answer 0 views 0

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    2021-10-04T15:39:33+00:00

    let one number be x
    Other number will be x-5
    193=(x)2 +(x-5)2
    193=x2+x2+25-10x
    193-25=2×2-10x
    168=2×2-10x
    2×2-10x-168=0
    Dividing by 2 on both sides
    x2-5x-84=0
    x2-5x+(5/2)^2-(5/2)^2-84=0
    (x+5/2)^2=(5/2)^2+84
    (x+5/2)^2=25/4+336/4
    (x+5/2)^2=25+336/4
    (x+5/2)^2=361/4
    x+5/2=+-rt361/2
    Then solve it equating both possibilities that is rt 361 as + or – then u will get alpha and beta. Plz mark it as brainliest I have worked hard so much to type.

    Plzzzzzzz……………

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