Question

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

Answer 1

let odd number be (2k+1)

Next odd number = (2k+3)

Sum of squares = (2k+1)^2 + (2k+3)^2 = 8k^2+16k+10

This sum is given as 290

8k^2 + 16k + 10 = 290

8k^2 + 16k - 280 = 0

8k^2 + 56k - 40k - 280 = 0

(8k-40) (k+7) = 0

k= 5 or k= -7

Since we want positive integers ,

k= -7 is not possible

Therefore k = 5

Required consecutive odd numbers are 11 & 13

Answer 2

Step-by-step explanation:

Hence,

Let the first integer = x

Second Integer = x+2

Also, given that

Sum of the squares of both the numbers = 290

( First number )2 + ( Second Number ) 2 = 290

(x)2 + ( x+2 )2 = 290

Using ( a+b )2 = a2 + b2 + 2ab

x2 + x2 + 4 + 4x =290

2x2 + 4x + 4 - 290 = 0

mene jitne bhi 2 likhe bracket aur x ke baad vo sab variable ke uper side likhne hai

Hope you understand it.