find three consecutive positive odd integers 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

Let the numbers be 2n-1, 2n+1, & 2n+3

Then (2n-1)² + (2n+1)² = (2n+3)³ + 65

Expand these terms and collect to get 4n² - 12n - 72 = 0

This has two real integer roots, each of which gives a set of 3 odd integers which satisfy the requirements of the problem.

Use the quadratic formula to get the roots.

Answer 2

The main trick here is to express all three integers in terms of one variable. Consecutive odd (or even) integers are separated by 2. The numbers can be (n), (n+2), and (n+4).

Now we can use the description to write an expression that can be solved for n.

The sum of the squares of the first two numbers: n² + (n+2)²

Is (=) 65 more than the square of the third (n+4)² + 65.

n² + (n+2)² = (n+4)² + 65

Now solve this for n.

n² + n² + 4n + 4 = n² + 8n + 16 + 65

2n² -n² + 4n - 8n + 4 - 16 - 65 = 0

n² - 4n -77 = 0

(n-11)(n+7) = 0

n = 7, -11

The solutions must be positive integers, so:

n=7

The answers are 7, 9, 11

Answer 3

Hi Maha!

Let's write this situation out with numbers. It might be a little clearer to see this way.

Starting with the first number in the sequence, we can just call it x.

The next number is the next odd integer. Odd numbers are always 2 apart from each other (think 3, 5, 7...).. so to describe the second number, we can express it as x + 2. We know it has to be 2 more than the first number.

The third number is the next odd integer, so we can think of this as x + 2 + 2 because it has to be 2 more than the second number. Let's simplify it to be x + 4.

Now we have expressions for our three mystery numbers: x, x + 2, and x + 4. We just have to figure out what x is in order to answer the actual question and figure out what the numbers actually are. You are given additional information: the sum of the squares of the first two numbers is greater than the square of the third by 65. This can be expressed in an equation as:

x² + (x + 2)² = (x + 4)² + 65

Let's go through and simplify by expanding our brackets and combining like terms:

x² + x² + 4x + 4 = x² + 8x + 16 + 65

2x² + 4x + 4 = x² + 8x + 81

x² - 4x - 77 = 0

Now we have a trinomial. In order to solve for x, we need to factorize it. Think about the factors of -77 that can combine to create -4. The numbers are -11 and 7.

(x - 11)(x + 7) = 0

x = 11, x = -7

Two possibilities, but only one is correct. Look at the first sentence in the problem. It says POSITIVE integers. Therefore, x cannot be -7. Instead it has to be 11. Which means the three numbers are 11, 13, and 15! Let's check our work to be 100% sure.

11² + 13² = 15² + 65

121 + 169 = 225 + 65

290 = 290

Hope this helped! I kind of skimmed through the explanation of the actual algebra involved with solving this equation. If you need a more in-depth explanation, let me know! :)