Find four consecutive natural numbers if the sum of the first and third numbers is 5 times less than

Find four consecutive natural numbers if the sum of the first and third numbers is 5 times less than the product of the second and fourth numbers.

Let the first of four consecutive natural numbers be x, then the second number is (x + 1), the third number is (x + 2), the fourth number is (x + 3). The sum of the first and third numbers is x + x + 2 = 2x + 2. The product of the second and fourth numbers is (x + 1) (x + 3). By the condition of the problem, it is known that the sum of the first and third numbers is 5 times less than the product of the second and fourth numbers. To equalize these values, you need to multiply the sum of the first and third numbers by 5, i.e. this will be 5 (2x + 2) and this value will be (x + 1) (x + 3). Let’s make an equation and solve it.

5 (2x + 2) = (x + 1) (x + 3);

10x + 10 = x ^ 2 + 3x + x + 3;

x ^ 2 + 3x + x + 3 – 10x – 10 = 0;

x ^ 2 – 6x – 7 = 0;

D = b ^ 2 – 4ac;

D = (- 6) ^ 2 – 4 * 1 * (- 7) = 36 + 28 = 64; √D = 8;

x = (- b ± √D) / (2a);

x1 = (6 + 8) / 2 = 14/2 = 7 – the first number;

x2 = (6 – 8) / 2 = – 2/2 = – 1 is not a natural number;

x + 1 = 7 + 1 = 8 – the second number;

x + 2 = 7 + 2 = 9 – the third number;

x + 3 = 7 + 3 = 10 – the fourth number.

Answer. 7, 8, 9, 10.