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