### Video Transcript

Find the solution set of the equation 18 over π₯ squared plus five over π₯ equals one, giving values to three decimal places.

One of the things we might realize here is that our variables of π₯ squared and π₯ are both given on the denominators. So letβs see if we can write it in the general form of a quadratic equation, ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π are constants and π is not equal to zero. If π was equal to zero, weβd have a linear equation. We can begin by multiplying both sides of our equation by π₯ squared, which gives us 18 plus five π₯ equals π₯ squared.

In order to get all of our terms on one side of the equation equal to zero, then weβll need to subtract five π₯ and subtract 18 from both sides, which leaves us with zero equals π₯ squared minus five π₯ minus 18, which can be rewritten as π₯ squared minus five π₯ minus 18 equals zero. The question asks us to solve or find the solution set of this equation, which means finding all the values of π₯ which make this equation true. We can solve a quadratic equation in a number of different ways, for example, through factoring. However, we are asked to give our solution set to three decimal places, which is usually a bit of a clue that we canβt factor. We could therefore alternatively solve this by completing the square or by using the quadratic formula. But in this video, weβll focus on the quadratic formula.

The quadratic formula tells us that if we have a quadratic equation in the form ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π are constants and π is not equal to zero, then π₯ is equal to negative π plus or minus the square root of π squared minus four ππ over two π. As we have our quadratic equation in this form, then we can take the values of π, π, and π and fill those into the quadratic formula. The value of π is the coefficient of π₯ squared, which is one. The value of π is the coefficient of π₯, which is negative five. And the value of π will be negative 18. Itβs always worthwhile writing these values of a, π, and π down clearly in our workings so that when weβre using the formula, we can easily refer to them.

So we take the formula and fill in our values of π, π, and π. We need to be really careful when weβre using the quadratic formula when we have negative values. So we have π₯ is equal to negative negative five plus or minus the square root of negative five squared minus four times one times negative 18 over two times one. When we simplify this, we have π₯ is equal to five plus or minus the square root of 25 plus 72 over two. 25 plus 72 simplifies to 97, which leaves us with π₯ is equal to five plus or minus the square root of 97 over two. Now we need to remember that this plus or minus symbol indicates two different solutions: π₯ is equal to five plus the square root of 97 over two and π₯ is equal to five minus the square root of 97 over two.

At this point, we can pick up our calculators to find the decimal solutions for each of these calculations. Our first answer is that π₯ is equal to 7.4244 and so on. And as weβre asked for a value to three decimal places, then this means that we check our fourth decimal digit to see if itβs five or more. And as it isnβt, then our value is π₯ is equal to 7.424. The second solution rounds from π₯ is equal to negative 2.4244 and so on to π₯ is equal to negative 2.424 to three decimal places. As we were asked for a solution set, then we must give our answer with set notation. So our answer is that the solution set of 18 over π₯ squared plus five over π₯ equals one is the set 7.424, negative 2.424.