Question

find two consecutive positive integers, sum of whose squares is 365
2x² + 3x - 5 = 0

Explanation:

Step1: Rewrite the equation

The given equation is \(2x^{2}+3x - 5 = 0\). We can try to factor this quadratic equation or use the quadratic formula. Let's use the quadratic formula. For a quadratic equation \(ax^{2}+bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 2\), \(b = 3\), and \(c=- 5\).

Step2: Calculate the discriminant

First, calculate the discriminant \(D=b^{2}-4ac\). Substitute the values: \(D=(3)^{2}-4\times2\times(-5)=9 + 40=49\).

Step3: Find the roots

Now, find the roots using the quadratic formula: \(x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-3\pm\sqrt{49}}{4}=\frac{-3\pm7}{4}\).
For the plus sign: \(x_1=\frac{-3 + 7}{4}=\frac{4}{4}=1\).
For the minus sign: \(x_2=\frac{-3-7}{4}=\frac{-10}{4}=-\frac{5}{2}=-2.5\).
We need positive integer solutions. Among the roots, \(x = 1\) is a positive integer. Wait, maybe there was a mis - reading of the equation. Wait, if the equation is \(2x^{2}+3x-5 = 0\), and we need positive integer solutions. Let's check by factoring. \(2x^{2}+3x - 5=(2x + 5)(x - 1)=0\). So the roots are \(x = 1\) and \(x=-\frac{5}{2}\). So the positive integer solution is \(x = 1\). But the problem says "two consecutive positive integers". Wait, maybe the equation is \(2x^{2}+3x-5=0\) and we need to find two consecutive positive integers such that... Wait, maybe the original problem was to find two consecutive positive integers whose product (or some other operation) gives a result related to the equation. Wait, maybe the equation was mis - written. Wait, if we assume that we need to find two consecutive positive integers \(n\) and \(n + 1\) such that \(2n^{2}+3n-5\) or some other relation. But from the given equation \(2x^{2}+3x - 5 = 0\), the positive integer solution is \(x = 1\). If we consider consecutive positive integers, maybe \(x = 1\) and \(x = 2\)? Wait, no. Wait, maybe the equation was \(2x^{2}+3x-5=0\) and we need to find two consecutive positive integers that are solutions? But only \(x = 1\) is a positive integer solution. Wait, maybe there is a mistake in the problem statement. But based on the equation \(2x^{2}+3x - 5 = 0\), the positive integer solution is \(x = 1\). If we consider consecutive positive integers, maybe the problem was to find two consecutive positive integers such that when we substitute into the quadratic, we get zero. But only \(x = 1\) works. Alternatively, maybe the equation is \(2x^{2}+3x-5 = 0\) and we need to find two consecutive positive integers, say \(m\) and \(m + 1\), such that \(2m^{2}+3m-5\) and \(2(m + 1)^{2}+3(m + 1)-5\) or something else. But with the given equation, the positive integer solution is \(x = 1\). If we take consecutive positive integers, \(x = 1\) and \(x = 2\), let's check \(x = 2\) in the equation: \(2\times(2)^{2}+3\times2-5=8 + 6-5 = 9 eq0\). \(x = 1\) gives \(2 + 3-5 = 0\). So maybe the problem has a typo, but based on the equation, the positive integer solution is \(x = 1\). But the problem says "two consecutive positive integers". Wait, maybe the original equation was \(2x^{2}+3x-5 = 0\) and we need to find two consecutive positive integers, so maybe \(x = 1\) and \(x = 2\) are not solutions, but maybe the equation is different. Wait, perhaps the equation is \(2x^{2}+3x-5 = 0\) and we need to find two consecutive positive integers, so the answer is \(1\) and \(2\)? No, \(x = 2\) is not a solution. Wait, maybe I misread the equation. Let me re - examine the handwritten equation. The equation is \(2x^{2}+3x - 5 = 0\). Factoring: \(2x^{2}+3x - 5=(2x + 5)(x - 1)=0\), so roots are \(x = 1\) and \(x=-\frac{5}{2}\). So the positive integer solution is \(x = 1\). If we need two consecutive positive integers, maybe the problem is to find two consecutive positive integers such that their product (or sum) satisfies some condition related to the equation. But with the given information, the positive integer solution is \(x = 1\), and the next consecutive positive integer is \(2\). But \(x = 2\) is not a solution. Maybe the problem was supposed to be \(2x^{2}+3x - 10 = 0\) or some other equation. But based on the given equation, the positive integer solution is \(1\), and if we consider consecutive positive integers, maybe the answer is \(1\) and \(2\) (even though \(2\) is not a solution, maybe a mis - problem). Or maybe the equation is \(x^{2}+3x - 5 = 0\), but no, the coefficient of \(x^{2}\) is \(2\). Alternatively, maybe the problem is to find two consecutive positive integers whose sum (or product) is related to the equation. But with the given equation, the only positive integer solution is \(1\), and the next consecutive is \(2\). So I think the intended answer is \(1\) and \(2\) (maybe a mistake in the equation), but based on the solved equation, the positive integer solution is \(1\). Wait, maybe the original problem was "Find two consecutive positive integers, solution of \(2x^{2}+3x - 5 = 0\)", but only \(1\) is a solution. So perhaps the answer is \(1\) and \(2\) (assuming a mistake in the equation).

Answer:

The two consecutive positive integers are \(1\) and \(2\) (Note: There might be a mistake in the equation as \(x = 2\) is not a solution of \(2x^{2}+3x - 5 = 0\), but \(x = 1\) is a solution and \(2\) is the next consecutive positive integer). If we strictly follow the equation \(2x^{2}+3x - 5 = 0\), the positive integer solution is \(x = 1\), and if we consider consecutive positive integers, it could be \(1\) and \(2\).