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

Question

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

x^(2/5) + x^(1/5) - 6 = 0

Accepted Answer

Welcome back. I'm so glad you're here. We've got this equation X to the 2/ plus X to the 1/7 minus two equals zero. And we are to solve for X. Using the substitution method. Well, what is the substitution method? The substitution method is when you've got something that looks like it could almost be factored but not quite. And then you notice that the first two terms look very similar and that the exponents of the first term are double that of the exponents of the second term. Well, what you're going to do is you're going to take that second term. So in this case X raised to the 1/7 and you're going to set it equal to you can choose whatever variable you like. I'm going to use W. And anywhere you have an X to the 1/7 you're going to substitute a W. Well what good does that do you for the first term? We'll recall if you took X to the 1/7 and raised it to the power of two and what you do to one side, you do to the other. So now you have W squared on the right hand side, you multiply those exponents and you now have X to the 2/7 equals W squared and X to the 2/7. That's exactly what our first term is. So now instead of X to the to seventh, it's now W squared plus W minus two equals zero. And that looks much more easily. Factory ble. So let's go ahead and do that. We'll set up our parentheses equals zero. So what times what equals W squared? That's going to be a W. Times of W and what times what equals negative two. Well it's going to be a positive and a negative and it has to add up to equal positive one. So let's do a positive two minus one. And we can always go back and foil this really quick to make sure that we chose are factors correctly didn't inadvertently make a positive negative and negative positive somewhere. So first W times W is W squared our outsides W times negative one is negative one. W insides two times W is plus two. W. And our last two times negative one is negative two. We can combine our like terms in the middle and we've got W squared minus W plus two, W is plus one. W and bring down that minus two. And that does match our previous equation. We chose our factors correctly. Now we can take those two factors W plus two and W minus one and set them both equal to zero. To solve for W. For W plus two equals zero will subtract two from both sides. So we have W equals negative two and for w minus one equals zero. We'll add one to both sides and we have W equals one. And before you rejoice because we're very close to getting it. We need to recall that W isn't what we're looking for. We're trying to solve for X. So anywhere we have a W. We can substitute X. Raised to the 1/7. So instead of W. Equals negative two, it's X. To the 1/7 equals negative two. And instead of W equals one it's X. To the 1/7 equals one. And how are we going to get rid of that? To the 1/7? Well we'll raise it to a higher exponent will raise it to the seventh. And what we do to one side we have to do to the other. So now X. To the 1/7 to the seventh. That's just X. To the first. So X equals a negative two to the seventh is negative 128. Doing the same thing on the other side here, raising it to the seventh power. So we've just got an X. On the left hand side but now one raised to the seventh, that's just one. So this one is X equals one, and X equals negative 128. We look at our answer choices and that matches 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^(2/5) + x^(1/5) - 6 = 0

Key Concepts

Substitution Method

Quadratic Equations

Exponents and Radicals

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