The sum of the squares of two consecutive odd numbers is 970. find type integers

21, 23 and -21, 23

Step-by-step explanation:

The difference between two consecutive odd numbers is 2.

Let the smaller number be x. Then, the larger number is x + 2.

Now add their squares and set it equal to 970. Then  solve for x.

smaller number: x

larger number: x + 2

sum of the squares of the two numbers: x^2 + (x + 2)^2

equation: x^2 + (x + 2)^2 = 970

Solution of the equation:

x^2 + (x + 2)^2 = 970

Expand the square of the binomial:

x^2 + x^2 + 4x + 4 = 970

Combine like terms on the left side:

2x^2 + 4x + 4 = 970

Subtract 970 from both sides:

2x^2 + 4x - 966 = 0

Divide both sides by 2:

x^2 + 2x - 483 = 0

Now we need to factor the binomial. We need two number whose sum is 2 and whose product is -483. Look at 483. It is not divisible by 2, but it is divisible by 3.

483/3 = 161

161 is divisible by 7.

161/7 = 23

23 is prime, so the prime factorization of 483 is

483 = 3 * 7 * 23

Notice that 3 * 7 = 21, so

483 = 21 * 23

Use -21 and 23:

-21 + 23 = 2

-21 * 23 = -483

x^2 + 2x - 483 factors into (x - 21)(x + 23)

Now we continue solving the equation.

x^2 + 2x - 483 = 0

(x - 21)(x + 23) = 0

x - 21 = 0 or x + 23 = 0

x = 21 or x = -23

We let x equal the smaller of the two integers, so now we add 2 to find the second integer.

x = 21; x + 2 = 23

x = -23; x + 2 = -21

There are two sets of consecutive odd numbers that satisfy this problem.

21, 23

-23, -21

The problem did not state positive or negative consecutive odd numbers, so both solutions are valid.

the first option is the answer.