The sum of the squares of two natural consecu | vedclass

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Question

(D) Let the two natural consecutive odd numbers be $n$ and $(n+2)$.According to the question,$n^{2} + (n+2)^{2} = 394$$n^{2} + n^{2} + 4 + 4n = 394$$2

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) Let the two natural consecutive odd numbers be $n$ and $(n+2)$.According to the question,$n^{2} + (n+2)^{2} = 394$$n^{2} + n^{2} + 4 + 4n = 394$$2n^{2} + 4n - 390 = 0$Dividing by $2$:$n^{2} + 2n - 195 = 0$Factoring the quadratic equation:$n^{2} + 15n - 13n - 195 = 0$$n(n+15) - 13(n+15) = 0$$(n-13)(n+15) = 0$Since $n$ must be a natural number,$n = 13$.The two numbers are $13$ and $15$.The sum of the numbers is $13 + 15 = 28$.

The sum of the squares of two natural consecutive odd numbers is $394$. The sum of the numbers is

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