Solve for x: x^(5/6) + x^(2/3) - 2x^(1/2) = 0

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Accepted Answer

Welcome back. I'm so glad you're here. We've got this beautiful equation here, we have four X. To the 5/6 plus two X to the two thirds minus six X to the one half equals zero. Well normally when we're solving for X we want something that's a little bit more like a quadratic equation. Like a X squared plus bx plus C equals zero. And this is it's close to that. There's three terms on the left but this third term isn't a constant. So let's see if we can get it to be a constant. Let's see if we can factor out that X to the one half and any other constant term that we could factor out. So looking at all three, we could try factoring out a two X to the one half. And let's do that division off to the side here, that'll be four X. To the 56 divided by two X. To the one half. Let's do four divided by two. That gives us two and X to the 5/6 divided by X. To the one half. We're going to subtract those exponents when we're dividing and so we are going to have a 3/6. I sound like a snake. A 3/6 there in the denominator. So subtracting our denominators leaves us with X to the 5/6 minus 36 is or in other words two X to the one third. Yes. Get rid of that sticks in the denominator. Thank you very much. Okay so putting that up here for the first term inside the parentheses is two X To the one third. Okay next 12 X. To the two thirds divided by two X. To the one half. The two divided by two. That's just one great. And X to the two thirds divided by X. To the one half those need a common denominator to subtract the exponents. And oh no two thirds is three or 4/6. Excuse me. Two thirds is 4/6 and one half is 3/6 so 4/6 minus 3/6 gives us X to the 1/6. Okay go back with our sixths. So two X to the one third plus X to the 1/6. Now that third term take a negative six X to the one half divided by two X. To the one half. Well negative six divided by two is negative three. And we know that that X to the one half divided by X. To the one half that cancels out which is what we designed it to do so X to the 16 minus three. To finish up those parentheses here equals zero. Great. Except what's in here. Well those exponents are still a bit wonky but at least they're just in the first two terms. And we can see if we can use substitution to use substitution here. You recall from previous lessons that we're just going to pick any variable other than X. And we're going to take the middle terms variable and its exponents. So X. To the 1/6. And we're going to set that equal to, I'm just going to choose W as my variable. So um we substitute W. For the middle term. And what about this? X to the one third? Well if we take X to the 1/ and we square it, what we do to one side we do to the other and X to the 1/6 to the second power probably X to the 2/6. Or in other words X to the one third equals W squared and X to the one third. That's exactly what we want to replace. So this does work for substitution. We can bring down the rest two X to the one half times parentheses, two W squared plus W minus three. End parentheses equals zero. Beautiful look at this. We can factor this here in the middle. Okay, what times what gives us two W squared? One of them will be a two W and one will be a W. And what? Two factors when multiplied equal? A negative 31 will be positive, one will be negative. And then when foiled out with that to W. And the W add up to be a positive one. W Well in this case that's going to be a positive three and a negative one. And we can foil that out just to double check. So two W times W. It's going to be two W squared. Our outsides, two W times negative one is negative two wr insides. Three W. Three times W is plus three W. And our last three times negative one is minus three, combine our inside like terms we've got two W squared plus W minus three. Yes that matches what we had up here. So we factored correctly and that leaves us with three factors each of which we can set equal to zero. So we've got two X to the one half. We can set equal to zero. Now look at this two W plus three equals zero. Remember that is A W. And then we've got W minus one equals zero. And keep in mind that that is a W. But first we can sell for X. So we divide both sides by two. Okay, well that's X to the one half still equals zero and we can square both sides and okay, that's still X equals zero. So one of our solutions X equals zero. Now for two W plus three equals zero. First I'm going to sell for W. And then substitute in. So let's subtract three from both sides and we've got two W equals negative three, divide both sides by two. And we have W equals negative three halves, scroll down a little bit. And on this other one, W minus one equals zero will add one to both sides and we'll have a W equals one. But now we have to go back and solve for X. So anywhere there's a W. we're going to substitute back in an X to the 1/6. So here instead of W equals negative three halves, it's X to the 1/6 equals negative three halves. And for W equals one it's going to be X to the 1/6 equals one. Thought it was done with the sixths. Huh? Okay, solving for X, we're going to take both sides and raise it to the sixth power. And I bet you didn't see this solution coming for X to the 16 to the sixth. That's just X. And negative three halves to the sixth Power is 64th. Alright, Now for this third one will raise both sides to the 6th. What am I doing to the 6th power and X to the one. Sixth to the sixth is just X. And one of six. That's just one. Okay, there's our other X solutions, we have three of them. Yay, this is exciting. Except when we started We did not have a power of three for our X variable. We headed to the 5/6. So it is entirely possible that going through all this work when we pulled out an X here and then we artificially created these two ws for to excess. It's possible we created an extraneous solution. So we have to check our answers. So I'm going to pull this original equation down off the screen for a second. So that when I scroll down we can see all three of these solutions and the original equation. So now let's go ahead and start substituting, let's start with X equals zero. We've got four times zero raised to the 5/6 plus two times zero, raised to the two thirds minus six times zero, raised to the one half. And does that equal zero? Well, yeah, zero plus zero minus zero equals zero. We've got a winner, X does equal zero. And now for a slightly less beautiful um substitution here, we're gonna put 729 64th in for X. So four times 64th to the plus two times 729 64th to the two thirds minus six times 729 64th to the one half equals zero. Okay, I'm not going to do this every single step but I will break it down a bit because as you're putting this in the calculator, there could be a place where you mistyped something so hopefully seeing it broken down a bit, will let you see if you made an error somewhere where it was. So 729 2 divided by 64 raised to the 56. Power is 243 30 seconds plus two times 729 64th raised to the two thirds Power is 81 16th minus six times 700. 29 64th. Race to the one half, Power is 27 8th equals zero And doing that multiplication four times 243 30 seconds is 243 8th plus two times 81 16th is 8th minus six times 27 8th is 162 8th equal zero. In doing that. Addition and subtraction. You've got a common denominator. Now you are going to get 8th which does not equal zero, which tells us that this was an extraneous solution that we created when we were trying to solve for X. It happens. All right now let's test. X equals one. A little bit easier here. Alright, four times one raised to the 5/6 plus two times one raised to the two thirds minus six times one, raised to the one half equal zero. Well, one raised to any power is just going to give us one here. 562 3rd and one half is going to be one. So this is just four times one plus two times one minus six times one equals zero. Four times one is four plus two times one is two minus six times one is six equals zero. And yeah, four plus 26 minus six is zero. So this one does work. X equals one is a solid solution. So now let's head back up to the top. We look at our answer choices and one and 0 matches with answer choice. A well done, we'll catch you on the next one.

Solve for x: x^(5/6) + x^(2/3) - 2x^(1/2) = 0

Key Concepts

Exponents and Fractional Powers

Combining Like Terms

Factoring Polynomials

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