Question

Найди два последовательных натуральных числа, если известно, что сумма их квадратов на 43 больше произведения этих чисел. Запиши в поле ответа числа в порядке возрастания без пробелов и запятых​

Answer 1

Объяснение:

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Answer 2

Problem Analysis

We are given that the sum of the squares of two consecutive natural numbers is 43 more than their product. We need to find these two numbers.

Solution

Let's assume the two consecutive natural numbers are x and x+1.

According to the problem statement, the sum of their squares is 43 more than their product. We can write this as an equation:

x^2 + (x+1)^2 = x(x+1) + 43

Simplifying the equation:

x^2 + x^2 + 2x + 1 = x^2 + x + 43

Combining like terms:

2x^2 + 2x + 1 = x^2 + x + 43

Rearranging the equation:

x^2 + x - 42 = 0

Now we can solve this quadratic equation to find the values of x.

Quadratic Equation Solution

We can solve the quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 1, b = 1, and c = -42.

Substituting the values into the quadratic formula:

x = (-(1) ± √((1)^2 - 4(1)(-42))) / (2(1))

Simplifying the equation:

x = (-1 ± √(1 + 168)) / 2

x = (-1 ± √169) / 2

x = (-1 ± 13) / 2

We have two possible solutions for x:

1. x = (-1 + 13) / 2 = 12 / 2 = 6 2. x = (-1 - 13) / 2 = -14 / 2 = -7

Since we are looking for natural numbers, we can discard the negative value of x.

Therefore, the two consecutive natural numbers are 6 and 7.

Answer

The two consecutive natural numbers are 6 and 7.