# Video: Solving Rational Equations Involving Quadratic Equations

Determine the solution set of the equation β(8 / (π₯Β² β 12π₯ + 27)) β ((96π₯ β 216) / (β27π₯Β² + 12π₯Β³ β π₯β΄)) = β200.

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### Video Transcript

Determine the solution set of the equation. Negative eight over π₯ squared minus 12π₯ plus 27 minus 96π₯ minus 216 over negative 27π₯ squared plus 12π₯ to the third power minus π₯ to the fourth power equals negative 200.

So to answer this question, we can see that we have at least one quadratic. And we also have at least two other terms that can be factored. In an exam question like this, weβll always check if we can factor any expressions. And then potentially cancel any equivalent expressions that appear on the numerator and denominator. Weβll also have to pay careful attention to the negative signs that appear in this question.

We can start by noticing that, on the denominator of our second term, we have π₯ squared, π₯ to the third power, and π₯ to the fourth power. Which means that we can take out π₯ squared as a common factor. In fact, as we begin to take out π₯ squared, we start to notice that our coefficients of our remaining terms are very similar to those given in the first fraction. The difference here though is that each coefficient has the opposite sign.

So instead of taking π₯ squared as a common factor, letβs try negative π₯ squared. The first term in our parentheses is found by considering what we multiply negative π₯ squared by to get negative 27π₯ squared, which would be 27. The second term in parentheses is found by calculating 12π₯ to the third power divided by negative π₯ squared, which is negative 12π₯. And for our third term, we divide negative π₯ to the fourth power by negative π₯ squared, which will give us plus π₯ squared for the final term.

We can now see that we have the same terms in both denominators. So letβs rewrite the second denominator so that it matches the first one. So now we have very similar denominators of our fractions. Weβre a step closer to solving this equation.

Before we look at the fraction arithmetic, we can see that our second term is subtracted and the entire denominator is multiplied by a negative sign. This will be the equivalent of adding the entire second term with a positive denominator. So next, in order to add fractions, we need to have two fractions that have the same denominators. We can achieve this by multiplying our entire first fraction by π₯ squared on the numerator and π₯ squared on the denominator.

We will now keep our denominators as they are and simply add the fractions by adding the numerators. We can write this negative sign by the eight π₯ squared, creating negative eight π₯ squared. Because we want to ensure that it doesnβt also include any other terms that we add on.

Our equation now becomes negative eight π₯ squared plus 96π₯ minus 216 over π₯ squared times π₯ squared minus 12π₯ plus 27 equals negative 200. As we consider the next steps to take, we might notice that we have an eight π₯ squared on our numerator. We might wonder then if our other coefficients on the numerator are also divisible by eight.

A quick check would tell us that 96π₯ divided by eight is 12π₯. We might then notice a familiar pattern, particularly when we observe that negative 216 divided by eight gives us negative 27. So again, we have similar coefficients with opposite signs. So letβs take our numerator and take out the common factor of negative eight.

Therefore, on our numerator, we have negative eight and then parenthesis. Our first term will be π₯ squared, since negative eight times π₯ squared gives us negative eight π₯ squared. Our second term in the parenthesis is negative 12π₯, since negative eight times negative 12π₯ gives us 96π₯. And our final term of 27 since negative eight times 27 gives us negative 216.

We have created two identical expressions, one on the numerator and one on the denominator. And these can effectively be cancelled out, leaving us with a much simpler equation. Negative eight over π₯ squared equals negative 200. We can cancel our negative signs from both sides of the equation, to give us eight over π₯ squared equals 200.

Multiplying both sides of our equation by π₯ squared, we have eight equals 200π₯ squared. We then divide both sides of our equation by 200 to get eight over 200 equals π₯ squared. And continuing to use a non-calculator method, we can simplify our fraction to one over 25 equals π₯ squared.

And finally, to find π₯, we take the square root of both sides of our equation, giving us that π₯ is equal to the square root of one over 25. We know that 25 is equal to five squared. So we have our value one-fifth. We also recognise that there are two values for the square root of our number, a positive value and a negative value. Therefore, π₯ equals one-fifth or negative one-fifth. And we write our answer as a solution set as required, one-fifth, negative one-fifth.