Question

Find two consecutive odd positive integers,sum of whose squares is 290.​

Answer 1

Two numbers are 11 and 13

Step-by-step explanation:

Let us consider two numbers are a and b.

According to your question

${a}^{2} + {b}^{2} = 290$

Now two numbers are consecutive positive odd numbers. So they are differ by 2 ( like 1 ,3). Then if a is greater number then a-b= 2 or a= 2+b. Put the value of a in the above equation,

$( {2 + b)}^{2} + {b}^{2} = 290 \\ {b}^{2} + 2b + 2 = 145 \\ {b}^{2} + 2b - 143 = 0 \\ b = 11$

Then a= 11+2=13.

So, the numbers are 11 and 13.