Find two positive whole number which differ by 5 and where the sum of their square is 193
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Hi,
Here is your 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.
Hope this helps you.
Here is your 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.
Hope this helps you.
Thanks for brainliest
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