2a+2ab+2b
I need a lot of help in this one.
it says find two consecutive positive integers such that the sum of their square is 85.
how would i do this one i have no clue i know what are positive integers.but i don't know how to figure this out.
Let n be a postive integer, then n+1 is the next integer.
sum the squares of those noumbers:
n^2 + n^2 + 2n +1=85
solve for n.
what is two consecutive positive integers square of 85
To find two consecutive positive integers whose sum of their squares is 85, you can follow these steps:
1. Let's assume the first positive integer is "n".
2. The second positive integer will then be "n + 1" since they are consecutive.
3. The sum of their squares would be: n^2 + (n + 1)^2.
4. Setting this sum equal to 85, we have the equation: n^2 + (n + 1)^2 = 85.
5. Expanding the equation, we get n^2 + n^2 + 2n + 1 = 85.
6. Simplifying further, we have 2n^2 + 2n + 1 = 85.
7. Rearranging the equation, we obtain 2n^2 + 2n - 84 = 0.
8. To solve this quadratic equation, we can either factor it or use the quadratic formula.
- Factoring: 2n^2 + 2n - 84 = 0 => (n - 6)(2n + 14) = 0.
So, either n - 6 = 0 or 2n + 14 = 0.
Solving each equation separately, we find that n = 6 or n = -7.
- Quadratic formula: n = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 2, b = 2, and c = -84.
Substituting these values into the quadratic formula, we get:
n = (-2 ± √(2^2 - 4*2*(-84))) / (2*2)
Simplifying further, we have n = (-2 ± √(4 + 672)) / 4
= (-2 ± √676) / 4
= (-2 ± 26) / 4
= 6 and -7.
9. Since we are looking for positive integers, the only valid solution is n = 6.
10. Therefore, the two consecutive positive integers whose sum of squares is 85 are 6 and 7.
So, the answer is 6 and 7.