I am not sure what this question is asking. Please help.
Find two consective postitve intergers such that the sum of their squares is 85.
n^2 + (n+1)^2 = 85
That equation can be rewritten
2 n^2 + 2n +1 = 85
or
n^2 + n - 42 = 0
This can be factored to obtain 2 solutions. Only one of them is a postive integer. That will be the lowest of the two numbers you want.
Sure! To find two consecutive positive integers such that the sum of their squares is 85, we can follow these steps:
1. Let's assume the first positive integer is "x".
2. Since we are looking for consecutive integers, the second positive integer will be "x + 1" (because it follows the first one).
3. The sum of their squares can be represented as: x^2 + (x + 1)^2.
4. We can simplify the equation: x^2 + (x + 1)^2 = 85.
5. Expanding the equation: x^2 + (x^2 + 2x + 1) = 85.
6. Combining like terms: 2x^2 + 2x + 1 = 85.
7. Moving all terms to one side of the equation: 2x^2 + 2x - 84 = 0.
8. Simplifying the equation further: x^2 + x - 42 = 0.
9. Now we need to factor the quadratic equation. To find two numbers whose product is -42 and whose sum is 1, we can rewrite the quadratic equation as: (x + 7)(x - 6) = 0.
10. Setting each factor equal to zero: x + 7 = 0 or x - 6 = 0.
11. Solving for "x" in each equation: x = -7 or x = 6.
12. Since we are looking for positive integers, we discard the negative solution (x = -7).
13. Therefore, the first positive integer is x = 6.
14. The second positive integer is x + 1 = 6 + 1 = 7.
So, the two consecutive positive integers whose sum of squares is 85 are 6 and 7.