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

Question

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

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Maya 3 weeks 2021-10-04T15:03:13+00:00 2 Answers 0 views 0

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    0
    2021-10-04T15:04:21+00:00

    The two numbers are 7 and 12.

    Step-by-step explanation:

    Let x and y be the two numbers.

    Then it is given that x = y + 5 ————(1)

    x² + y² = 193 ————-(2)

    Substituting (1) in (2), we get:

    (y + 5)² + y² = 193

    25 + y² + 10y + y ² = 193

    2y² + 10y – 168 = 0

    Let’s solve the quadratic equation.

    y² + 5y – 84 = 0

    y² + 12y – 7y – 84 = 0

    y (y + 12) – 7(y + 12)

    (y + 12) (y – 7) = 0

    Therefore y = -12 or +7

    It is mentioned that the numbers are positive whole numbers. So y can only be +7.

     Substituting y = 7 in equation 1, we get: x = 12

    So the two numbers are 7 and 12.

    0
    2021-10-04T15:05:05+00:00

    Answer:

    Let the two positive whole numbers be x and y.

    Given that these numbers differ by = 5

    Therefore x – y = 5 

    x = 5 + y

    Also given that the sum of their squares = 193

    Therefore

    x² + y² = 193

    Substituting the value of x

    (5 + y)² + y² = 193

    25 + y² + 10y + y ² = 193

    2y² + 10y – 168 = 0

    y² + 5y – 84 = 0

    y² + 12y – 7y – 84 = 0

    y(y + 12) – 7(y + 12)

    (y + 12) (y – 7) = 0

    y = -12 or y = 7

    Rejecting y = -12 as given they are positive whole numbers.

    Therefore y = 7

    x – 7 = 5

    x = 12

    The two numbers are 12 and 7.

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