Question

Three consecutive odd numbers are such that the sum of the squares of the first two
numbers is greater than the square of the third by 65. Find the numbers.​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

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.