Question

Two numbers have a geometric mean of \(10 .\) One number is 21 less than the other. Find the numbers.

Short answer

The numbers are 25 and 4.

Step-by-step solution

Formulate the equations based on given conditions

Let's denote the two numbers as \(x\) and \(y\). From the conditions we can form the following two equations: \n1) \(x*y = 10^2\) (since the geometric mean is the square root of the product of the numbers) \n2) \(x - y = 21\) (since one number is 21 less than the other)

Solve the system of equations

To simplify solving this system of equations, you can express \(y\) from the second equation: \(y = x - 21\). Now, substitute this value into the first equation: \(x*(x - 21) = 100\), which simplifies to \(x^2 - 21x - 100 = 0\). This is a quadratic equation and can be solved using the quadratic formula or factoring.

Find the roots of the quadratic equation

The roots of the equation \(x^2 - 21x - 100 = 0\) can be found by factoring: \((x - 25)(x + 4) = 0\). The solutions are \(x = 25\) and \(x = -4\). Since in our context \(x\) must be a positive number (it represents a number in our original problem), we discard \(x = -4\) as extraneous.

Find the value of y

We substitute \(x = 25\) into our second original equation \(x - y = 21\) to find \(y\). This gives us \(y = 25 - 21 = 4\).

Review the Solution

The solutions \(x = 25\) and \(y = 4\) satisfy both original conditions. They have the geometric mean of 10 and one number is 21 less than the other. Therefore, the two numbers we seek are 25 and 4.