consecutive odd integer problem

find two odd integers the sum of whose squares is 130.
(sq)=squared

let x= 1st odd int
let x+2=2nd odd int

(2x+2)(2x+2)=130
4x(sq) + 4x+ 4x +4=130
4x(sq) + 8x -126= 0

2(2x(sq)+4x-63)=0
My question: WHAT ARE THE TWO DARN #s!!!

The squares of the first few odd integers are: 9, 25, 49, 81, 121

Two pairs of these add up to 130, but the square roots of which two of these are consecutive?

the two numbers to factor the equation.

pair of #s that multiplies to -126
and the same pair that adds to a 4.

consecutive odd integer problem

find two odd integers the sum of whose squares is 130.
(sq)=squared

let x= 1st odd int
let x+2=2nd odd int

x^2 + (x + 2)^2 = 130
x^2 + x^2 + 4x + 4 = 130
2x^2 + 4x - 126 = 0
x^2 + 2x - 63 = 0

I bet you can find the answers now.

To find the two odd integers, we need to solve the quadratic equation 4x² + 8x - 126 = 0.

One way to solve this equation is by factoring. However, it seems difficult to find two numbers that multiply to -126 and add up to 8, so factoring might not work in this case.

Alternatively, we can use the quadratic formula to find the values of x. The quadratic formula is given by:

x = (-b ± √(b² - 4ac)) / 2a

In this case, a = 4, b = 8, and c = -126. Substituting these values into the quadratic formula, we get:

x = (-8 ± √(8² - 4 * 4 * -126)) / (2 * 4)

x = (-8 ± √(64 + 2016)) / 8

x = (-8 ± √2080) / 8

To simplify the calculation, let's find the square root of 2080:

√2080 ≈ 45.63

Now, we can substitute this value into the equation:

x = (-8 ± 45.63) / 8

For the first solution, we have:

x₁ = (-8 + 45.63) / 8

x₁ = 37.63 / 8

x₁ ≈ 4.70

Since we are looking for odd integers, x₁ is not an integer and is therefore not a valid solution.

For the second solution, we have:

x₂ = (-8 - 45.63) / 8

x₂ = -53.63 / 8

x₂ ≈ -6.70

Since we are looking for odd integers, x₂ is not an integer and is therefore not a valid solution either.

Therefore, there are no two odd integers whose sum of squares is 130.