Question

A positive real number is 1 less than another. When 2 times the larger is added to the square of the smaller, the result is 13. Find the numbers. (If applicable, write your answers in the form p±q√r) ​

Answer 1

Let x be the smaller number, then the larger number is x + 1.

From the problem, we know that:

2(x + 1) + x^2 = 13

Simplifying and rearranging:

x^2 + 2x - 11 = 0

Using the quadratic formula:

x = (-2 ± sqrt(2^2 - 4(1)(-11))) / (2(1))

x = (-2 ± sqrt(48)) / 2

x = (-2 ± 4sqrt(3)) / 2

x = -1 ± 2sqrt(3)

Since the problem states that the numbers are positive, we can ignore the negative solution. Therefore, the smaller number is:

x = -1 + 2sqrt(3)

And the larger number is:

x + 1 = 2sqrt(3)

So the two numbers are -1 + 2sqrt(3) and 2sqrt(3).