Find two consecutive odd integers such that the square of the larger is one less than twice the square of the smaller.
5 and 7
Check -- do those numbers work?
To find two consecutive odd integers, we can assume a variable to represent the smaller odd integer. Let's call it "x." Since they are consecutive, we know that the larger odd integer will be x + 2.
According to the problem, the square of the larger integer is one less than twice the square of the smaller integer. Mathematically, we can express this as:
(x + 2)^2 = 2x^2 - 1
Now, let's solve this equation step by step:
Expanding the left side of the equation:
x^2 + 4x + 4 = 2x^2 - 1
Rearranging the equation to bring all terms to one side:
2x^2 - x^2 - 4x - 4 + 1 = 0
Simplifying:
x^2 - 4x - 3 = 0
Now, we need to solve the quadratic equation:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = -4, and c = -3.
x = (-(-4) ± √((-4)^2 - 4(1)(-3))) / (2(1))
Simplifying:
x = (4 ± √(16 + 12)) / 2
x = (4 ± √28) / 2
x = (4 ± 2√7) / 2
x = 2 ± √7
So, the two consecutive odd integers are 2 + √7 and 2 - √7.
(n+2)^2 = 2n^2 - 1
You might think that it would be possible for n to be even, but since the right side must be odd, so must n^2 be odd, and thus any solutions you find will be odd.