The sum of the squares of two positive numbers is 233 and one number is 3 less than twice the other number. Find the numbers.
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The sum of the squares of two positive numbers is 233 and one number is 3 less than twice the other number. Find the numbers.
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Given question:-
The sum of the squares of two positive numbers is 233 and one number is 3 less than twice the other number. Find the numbers.
Solution:-
Given,
Sum of squares of two positive numbers = 233
One number + 3 = 2 × Other Number
According to the question,
~ Case | :-
x + 3 = 2y
x = 2y - 3 ......(1)
~Case || :-
[tex] {x}^{2} + {y}^{2} = 233.....(2)[/tex]
From equation, (1) and (2), we get,
[tex] {(2y - 3)}^{2} + {y}^{2} = 233[/tex]
[tex] {4y}^{2} - 9 - 12y + {y}^{2} = 233[/tex]
[tex] {5y}^{2} - 12y + 9 = 233[/tex]
[tex] {5y}^{2} - 12y + 9 - 233 = 0[/tex]
[tex] {5y}^{2} - 12y - 224 = 0[/tex]
[tex] {5y}^{2} - 40y + 28y- 224 = 0[/tex]
[tex]5y(y - 8) + 28(y - 8) = 0[/tex]
[tex](y - 8)(5y + 28) = 0[/tex]
[tex]y- 8 = 0 \\ y = 0 + 8 \\ y = 8[/tex]
Or
[tex]5y+ 28 = 0 \\ 5y = 0 - 28[/tex]
[tex]5y= - 28[/tex]
[tex]y = \frac{ - 28}{5} [/tex]
Negative value is not acceptable. Hence, y = 8.
x = 2(8) -3
x = 16 -3
x = 13.