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

Question

Solve each equation in Exercises 41–60 by making an appropriate substitution. x - 13√x + 40 = 0

Accepted Answer

Welcome back. I'm so glad you're here. We are to solve the following equation. We've got x minus 15 square root of x plus 54 equals zero. We're to solve it using the substitution method. Well, what does that mean? Well, when you've got this equation that looks like it almost could be factored in order to solve for X but not quite. And you look carefully at the first variable and the second variable and they're similar. And you notice that the exponents of the first one in this case X to the first is double that of the exponents of the second one. The squared of X. That's the same as X to the one half and double one half. That's one. So in this case we can use substitution. And how do we do that? We take that middle term the square root of X. And we're going to set that to any variable. We like we're going to substitute in this case a W for that square root of X. So now instead of negative 15 squared of X, that's negative 15. W. What about that first variable? Well, we know that if we were to take the squared of X and square it, what we do to one side we have to do to the other squared of X squared. That's just X. Which is what we have for our first variable and then that bring down that equals W squared. So we are going to substitute a W squared for the first variable. We bring down that minus 15 times the W and bring down the plus 54 equals zero. Great. Now we've got something that we can factor. So we set up our parentheses and we look at our first term here and we've got W squared. Well what gives us W square? That will be a W times W. And we look at our last term 54 and that has potentially lots of options. But most commonly we think of six and nine and hey, that'll get us to 15. But we notice that when they're added they need to equal a negative 15. So we'll go with a negative six times a negative nine to get 54. And if we ever need to, we can always just foil this really quick to double check and make sure we factored it correctly first, W times W is W squared. Our Outsides W times negative nine is negative nine. W insides negative six times W is negative six W. And our last negative six times negative nine is positive 54. And then we need to combine our like terms in the middle. Not factoring again, we go W squared minus nine W minus six. W gives us minus 15 W plus 54. And yes, this checks out that is the equation that we had previously. So we factored it correctly. Now we can take each of those two factors W minus six and W minus nine and set them equal to zero. So for w minus six, we're going to add six to both sides. So we have W equals six and for W minus nine, we're going to add nine to both sides and we have W equals nine. Now before you celebrate that we've got the answers, recall that we're not actually trying to find W. We are trying to find X. We need to sell for X. So anywhere we have a W we are now going to again substitute a square root of X. So instead of W equals six, it's the squared of X equals six and instead of W equals nine, it's the square root of X equals nine. Now, how do we undo that square root to get X by itself? We are going to square it and what we do to one side. We have to do the other. When we have the square root of X squared, that gives us x and six square that is 36. We're going to use the same method to get X by itself. With this squared of X equals nine. Will square both sides squared of X squared is X and nine squared is 81. So our two solutions for X are X equals 36 X equals 91 or 81. Excuse me? We look at our answer choices and that matches with answer choice B Well done. We'll catch you on the next one

Solve each equation in Exercises 41–60 by making an appropriate substitution. x - 13√x + 40 = 0

Key Concepts

Substitution Method

Quadratic Equations

Domain Restrictions for Square Roots

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