Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

Let's denote the number we are looking for as x.

The problem can be translated into the following equation:

x + 17 = 60/x

To solve this equation, we first multiply each side by x to get rid of the fraction:

x^2 + 17x = 60

Then, we rearrange the equation to set it equal to zero:

x^2 + 17x - 60 = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = 17, and c = -60.

We can solve this equation using the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a)

Substituting the values of a, b, and c into the formula, we get:

x = [-17 ± sqrt((17)^2 - 41(-60))] / (2*1) x = [-17 ± sqrt(289 + 240)] / 2 x = [-17 ± sqrt(529)] / 2 x = [-17 ± 23] / 2

This gives us two solutions: x = 3 and x = -20.

However, the problem asks for a positive number, so the number we are looking for is x = 3.

Sure, let's solve this step by step.

1. Let's denote the number we're looking for as x. According to the problem, we know that when this number is increased by 17, it equals 60 times the reciprocal of the number. We can write this as an equation: x + 17 = 60/x.

2. To solve for x, we first need to get rid of the fraction. We can do this by multiplying every term by x, which gives us: x^2 + 17x = 60.

3. Now, we can rearrange the equation to look like a quadratic equation: x^2 + 17x - 60 = 0.

4. We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. The factored form of the equation is (x - 3)(x + 20) = 0.

5. Setting each factor equal to zero gives the solutions x = 3 and x = -20.

6. However, the problem asks for a positive number, so the solution is x = 3.