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?
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!!!
3 answers
the two numbers to factor the equation.
pair of #s that multiplies to -126
and the same pair that adds to a 4.
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.
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.