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:
the two numbers to factor the equation.
pair of #s that multiplies to -126
and the same pair that adds to a 4.
found it set it up wrong was the problem :)
To factor the equation 4x^2 + 8x - 126 = 0, you need to find a pair of numbers that multiplies to -126 and adds up to 8.
First, list the factors of 126:
1 * 126 = 126
2 * 63 = 126
3 * 42 = 126
6 * 21 = 126
7 * 18 = 126
9 * 14 = 126
Now check the pairs to see which one adds up to 8:
1 + 126 = 127
2 + 63 = 65
3 + 42 = 45
6 + 21 = 27
7 + 18 = 25
9 + 14 = 23
The pair that adds up to 8 is 7 and 18.
So we can rewrite the equation as:
4x^2 + 8x - 126 = (2x + 7)(2x + 18) = 0
Now, you can set each factor equal to zero and solve for x:
2x + 7 = 0 or 2x + 18 = 0
2x = -7 2x = -18
x = -7/2 x = -18/2
Therefore, the two odd integers are x = -7/2 and x + 2 = -7/2 + 2 = 5/2.