three consecutive odd numbers are such that the sum of the squares of the first two numbers is greater than the square of third number by 65 .find the numbers
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Let the numbers be (2n+1),(2n+3) and (2n+5).
According to the Question,
(sum of the squares of the first two numbers) = (square of the third number) + 65
(2n+1)² + (2n+3)² = (2n+5)² + 65
(2n)²+2(2n)(1)+(1)² + (2n)²+2(2n)(3)+(3)² = (2n)²+2(2n)(5n)+(5)²+65
4n²+4n+1 + 4n²+12n+9 = 4n²+20n+25+65
8n²+16n+10 = 4n²+20n+90
4n²-4n-80 = 0
4(n²-n-20) = 0
n²-n-20 = 0
n²-5n+4n-20 = 0
n(n-5)+4(n-5) = 0
(n-5)(n+4) = 0
n=5 or n=-4
Putting the values of n=5,
2n+1=2(5)+1=11
2n+3=2(5)+3=13
2n+5=2(5)+5=15
Putting the values of n=-4,
2n+1=2(-4)+1=-7
2n+3=2(-4)+3=-5
2n+5=2(-4)+5=-3
So, the numbers are either 11,13&15 or -7,-5&-3 respectively.
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