Question

The two consecutive odd positive integers 'sum of whose square is 290

Answer 1

Answer:Here is your answer

Step-by-step explanation:

Answer 2

Positive Odd Integers are of the form - 2n - 1, 2n + 1, 2n + 3 and so on

ATQ,

$\begin{gathered}\sf (2n + 1)^2 + (2n - 1)^2 = 290 \\ \\ \longrightarrow \sf 4n^2 + 4n + 1 + 4n^2 - 4n + 1 = 290 \\ \\ \longrightarrow \sf 8n^2 = 288 \\ \\ \longrightarrow \sf n^2 = 36 \\ \\ \longrightarrow \sf n = 6 \ \ \ \{\because n \in Z^+\}\end{gathered}$

Now,

• 2n + 1 = 2(6)+1 = 13
• 2n - 1 = 2(6)-1 = 11

Two consecutive positive odd integers are 11 and 13.