The sum of the squares of three consecutive integers is 149. Find the integers.
I got 6,7,8 or -6,-7,-8
There's no reason why it would specifically be one or the other right? It could be either?
Thanks!!
correct
You're correct! The sum of the squares of three consecutive integers can indeed have two different solutions. In this case, both 6, 7, 8 and -6, -7, -8 are valid solutions.
To find these solutions, we can start by representing the three consecutive integers as n, n+1, and n+2.
The sum of their squares can be written as:
n^2 + (n+1)^2 + (n+2)^2 = 149
Expanding the squares and simplifying the equation, we get:
n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 = 149
Combining like terms, we have:
3n^2 + 6n + 5 = 149
Rearranging the equation, we get:
3n^2 + 6n - 144 = 0
Now we can solve this quadratic equation by factoring or using the quadratic formula. Factoring is a bit tricky here, so I'll use the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / (2a)
For the given equation, a = 3, b = 6, and c = -144. Substituting these values into the formula, we have:
n = (-6 ± √(6^2 - 4 * 3 * -144)) / (2 * 3)
Simplifying further, we get:
n = (-6 ± √(36 + 1728)) / 6
n = (-6 ± √1764) / 6
n = (-6 ± 42) / 6
Now we have two possible values for n:
1) n = (-6 + 42) / 6 = 36 / 6 = 6
2) n = (-6 - 42) / 6 = -48 / 6 = -8
Substituting these values back into the original equation, we can find the corresponding consecutive integers:
1) If n = 6, then the consecutive integers are: 6, 6+1=7, and 6+2=8
2) If n = -8, then the consecutive integers are: -8, -8+1=-7, and -8+2=-6
Both sets of consecutive integers satisfy the condition of the sum of their squares being 149. Therefore, your solutions of 6, 7, 8 and -6, -7, -8 are correct!