Solve each equation in Exercises 41–60 by making an appropriat...

Question

Solve each equation in Exercises 41–60 by making an appropriate substitution. ${(y - \frac{8}{y})}^{2} + 5 (y - \frac{8}{y}) - 14 = 0$

Accepted Answer

Welcome back. I'm so glad you're here. We have a doozy of a problem. Here it is X plus one over X. All squared minus 10 parentheses, X plus one over X again. And parentheses plus 16 equals zero. And we're asked to solve this Using the substitution method. Okay, when we are using the substitution method, we want to take a look at our second terms variable in this case that is X plus one over X. And we ask ourselves if we square our second terms variable. Does that give us our first terms variable? And in this case yes it does. The first terms variable is very clearly the square of the second terms variable. So great. We can continue with substitution. We take the second terms variable. So that is X plus one over X. And we set it equal to any variable variable. We like as long as it's not the one that we already have here. So I'm going to choose W anywhere. We have an X plus one over X. I'm going to substitute a W. Then we bring down the rest. So this is W squared minus 10. W plus 16 equals zero. And now we can factor this W squared minus 10. W plus 16. What times what gives us W squared? That's W times W. And what two numbers when multiply give us a positive 16. And when added give us a negative 10, that will be a negative eight and negative two. We have two factors and we can set each of them equal to zero. So w minus eight equals zero and w minus two equals zero. I am going to add eight to both sides of this first equation so that we have W equals eight. And I'm going to add two to both sides of the second equation and we get a W equals two, which is great. Except we're not solving for W we are solving for X. So anywhere there's a W I'm going to substitute an X plus one over X. So now it's X plus one over X equals eight and X plus one over X equals two. Let's get rid of that fraction in both equations will multiply every term times not two times X times our denominator to get rid of that fraction. So the first equation X times X will give us an X squared plus X times one over X will give us one X over X or one equals x times eight which is eight X. Second equation very similar will have X squared plus one equals two X. Now we've got quadratic equations and let's bring our eight X and R. Two X over to the left hand side so that our equations will be equal to zero. On the left will have X squared minus eight, X plus one equals zero. And on the right will have x squared minus two, X plus one equals zero. I'm going to look at that right hand equation for a moment here because this one can be beautifully factored what times what gives us X squared? That's X times x. What two numbers? When multiplied? Give us a positive one. And when added gives us a negative two, that's going to be a minus two. Excuse me a minus one and minus one. And we have two factors. Normally we set them both equal to zero but they're the same thing. So I'll just set X -1 equal to zero and we'll solve four. X -1 equals zero. And that will give us an X equals one. There is one of our solutions X equals one. All right now we've got to take a look at this X squared minus eight X plus one. We don't have two numbers that when multiplied give us a positive one. And when added give us a negative eight. So we're not going to be able to factor this one. This one we can pull out our good old friend the quadratic formula. If you recall from previous lessons, the quadratic formula is negative B plus or minus the square root of B squared minus four times a times C. All over two times A. Let's identify our A B and C. So the A. Is in front of the X squared. So A equals one. The B is in front of the xo B equals negative eight. The C is the constant term. So C equals one. Now I'm going to quickly fill those into our quadratic formula will have a negative B. So that's a negative negative eight plus or minus the square root of negative eight squared minus four times A. Which is one times C. Which is one all over two times one. Now simplifying we get negative negative eight is eight plus or minus. The square root of negative eight squared is 64 minus four times one times one is minus 4/2 times one which is two. We have eight plus or minus. The square root of 64 -4 is 60 Over to. Let's work on the square root of 60. Let's build a factor tree for 6060 is six times 10. Six is two times 3, 10 is two times five. So that's a two squared times three times five. That's 15. So two Squared Times 15. Let's fill that back in. We've got eight plus or minus the square root of two squared times 15. All over to and that is eight plus or minus that to square. That can come out of the radicals and that becomes too. And then leaving that square root of 15. All divided by two. We can divide those two terms in the numerator by 28 divided by two is four and two divided by two is one for the two squared 15 divided by two is one square to 15. So that simplifies to four plus or minus the square root of 15. I'm going to scroll back up and write that at the top four plus or minus the square to 15. Four which would give us four plus the square 2, 15 and four minus the squared of 15. Well let's now ask ourselves if it makes sense that we got three solutions for X. We look at the degree of our original equation and it's got this degree of two but it has an X and a one over X in here that are getting squared, those two are being added together and it's possible to have three solutions for X. It's also possible we came up with an extraneous solution. So let's go through and check each one. We're going to check one for plus the square to 15 and for minus the square to 15. And in order to do that, I'm going to bring down our original equation where we'll have a bit more room. Okay, So I'm gonna rewrite that x equals one and then we've also got this X equals four plus the square 2, 15 and four minus the square to 15. Well let's work on X equals one. First. A little bit easier to plug in there. Okay, so we've got parentheses one plus 1/ squared minus 10 times parentheses one plus 1/1. End, parentheses plus 16 equals zero, one plus 1/1. That's one plus one which is two. So we've got two squared minus 10 times. That's that same thing. Which is to plus 16 equals + squared is four minus 10 times two is 20 plus 16 equals +04 minus 20 is negative plus 16 does in fact equal zero. We've got a winner X does equal one. Now let's check the other two. We need to check X equals four plus the square to 15. And I mean if you want to go for it, go for it, trying to solve this algebraic lee but I highly recommend plugging this into a calculator and I'll show you what to plug into the calculator. You want a four A parentheses four plus the square to plus one divided by four plus the square to 15. End parentheses squared minus 10 times parentheses four plus the squirt of 15 plus 1/ plus the square to 15. End parentheses plus 16. You can plug that in, see if that equals zero. And while you're plugging that in, I'll show you how this other one looks. X equals for minus the square to 15. So we've got four minus the square to 15 plus 1/4 minus the square to 15, all squared minus 10 times four minus the square to 15 plus 1/4 minus the square to 15. And parentheses plus 16 equals zero. So if you need to pause here and if you didn't now continuing hopefully you've got that all plugged in. And yeah these two both do equal zero. So these solutions. All three of them work. We scroll back up to the top here and look at our answer choices and these three match answer choice, a well done. We'll catch you on the next one.

Solve each equation in Exercises 41–60 by making an appropriate substitution. ${(y - \frac{8}{y})}^{2} + 5 (y - \frac{8}{y}) - 14 = 0$

Key Concepts

Substitution Method

Quadratic Equations

Solving Rational Expressions

Watch next

Solve each equation in Exercises 41–60 by making an appropriate substitution. (y−8y)2+5(y−8y)−14=0\left( y - \frac{8}{y} \right)^2 + 5 \left( y - \frac{8}{y} \right) - 14 = 0

Key Concepts

Substitution Method

Quadratic Equations

Solving Rational Expressions

Watch next

Solve each equation in Exercises 41–60 by making an appropriate substitution. ${(y - \frac{8}{y})}^{2} + 5 (y - \frac{8}{y}) - 14 = 0$