### Video Transcript

The ratio π₯ to four equals four to π¦, so four is the geometric mean of π₯ and π¦. Find the geometric mean of π₯ plus one over π¦ and π¦ plus one over π₯.

Letβs begin by recalling the definition of a geometric mean. The geometric mean of two numbers π and π, which must have the same sign, is the square root of ππ. We can only find the geometric mean of two numbers which have the same sign because if the two numbers were of opposite signs, then the product would be negative. And the square root of a negative number gives a nonreal result. We are told that four is the geometric mean of π₯ and π¦, so we know that the square root of the product π₯π¦ is equal to four. Letβs also just explore this ratio for a moment. We are told that the ratio π₯ to four is the same as the ratio four to π¦.

If these two ratios are the same, then if we divide the value on the left of each ratio by the value on the right, we get the same result. So we have an equation, π₯ over four is equal to four over π¦. Multiplying both sides of this equation by the two denominators of four and π¦, we have π₯π¦ is equal to four squared. Then taking the positive square root of each side of this equation, we have that the square root of π₯π¦ is equal to four. And so we see that four is indeed the geometric mean of π₯ and π¦.

Now, we are asked to find the geometric mean of two other quantities: π₯ plus one over π¦ and π¦ plus one over π₯. Provided these two numbers are of the same sign then, their geometric mean is the square root of their product. Itβs the square root of π₯ plus one over π¦ multiplied by π¦ plus one over π₯. We can deduce that these two quantities do have the same sign because π₯ and π¦ have the same sign. If π₯ and π¦ are both positive, then every term involved in this expression is positive. And so weβre multiplying one positive value by another positive value. If, on the other hand, π₯ and π¦ are both negative, then one over π¦ and one over π₯ are also both negative. So everything in the expression is negative, and weβre multiplying two negative values together.

Letβs consider then how we can manipulate this expression, and weβll begin by distributing the parentheses. Using the FOIL method, we have π₯π¦ plus π₯ multiplied by one over π₯ plus one over π¦ multiplied by π¦ plus one over π₯ multiplied by one over π¦. Each of the terms in the center of this expansion simplify to one. And one over π₯ multiplied by one over π¦ is one over π₯π¦. So we have π₯π¦ plus one plus one plus one over π₯π¦, which simplifies to π₯π¦ plus two plus one over π₯π¦. The geometric mean of these two expressions then is the square root of π₯π¦ plus two plus one over π₯π¦.

Now next, we recall that we know the geometric mean of π₯ and π¦, the square root of π₯π¦ is equal to four. Squaring both sides of this equation tells us that π₯π¦ is equal to 16. And we can now substitute this value for π₯π¦ in two places in our expression for the geometric mean, which gives the square root of 16 plus two plus one over 16. Thatβs the square root of 18 and one sixteenth or the square root of 18.0625. And we can then use the calculator to evaluate this, and it gives 4.25.

So given that the geometric mean of π₯ and π¦ is four, we found that the geometric mean of π₯ plus one over π¦ and π¦ plus one over π₯ is 4.25.