Answered

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The difference of two positive integers is 5 and the sum of their squares is 433. Find their integers.

Sagot :

Hrishii

Answer:

12, 17

Step-by-step explanation:

Let the two positive integers be x and (5 + x).

According to the given description:

[tex] {x}^{2} + {(x + 5)}^{2} = 433 \\ {x}^{2} + {x}^{2} + 10x + 25 = 433 \\ \\ 2 {x}^{2} + 10x + 25 - 433 = 0 \\ \\ 2 {x}^{2} + 10x - 408 = 0 \\ \\ 2( {x}^{2} + 5x - 204) = 0 \\ \\ {x}^{2} + 5x - 204 = 0 \\ \\ {x}^{2} + 17x - 12x - 204 = 0 \\ \\ x(x + 17) - 12(x + 17) = 0 \\ \\ (x + 17)(x - 12) = 0 \\ \\ x + 17 = 0 \: or \: x - 12 = 0 \\ \\ x = - 17 \: or \: x = 12 \\ \\ \because \: x \: is \: a \: + ve \: integer \\ \\ \implies x \neq - 17 \\ \\ \therefore \: x = 12 \\ \\ 5 + x = 5 + 12 = 17 \\ [/tex]

Thus, the two positive integers are 12 and 17.

rysakapoor19

Answer:

12 and 17

Step-by-step explanation:

17 - 12 = 5, and 17^2 (289) + 12^2 (144) = 433

Hope this helps :D

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