In Exercises 91–100, find all values of x satisfying the given...

Question

In Exercises 91–100, find all values of x satisfying the given conditions. $y = x - \sqrt{x - 2} \quad ext{and} \quad y = 4$

Accepted Answer

Welcome back. I'm so glad you're here. We are Given this equation we have why equals two X minus the square root of four. X minus 24. And we're told that Y equals 24 we are to solve for X. Alright well let's substitute in for that Y instead of Y equals it'll be 24 equals and then bring down the rest 22. Ex excuse me. Two X minus the square root of four. X minus 24. Okay now let's look inside of this radical and let's see if we can simplify this a bit four X minus 24. Will those both have a four in them? So we could factor a four out. So before times X minus six and that's still the square root of and bring down our 24 equals two X minus the square root of four times x minus six. Now we recall from the product rule that when you have the square root of A times B, that's the same as the square root of a. Times the square root of B. So looking under that radical, we've got the square root of four times x minus six. Which is the same as the square root of four times the square root of x minus six. And we can take the square root of four. The square to four is two. So now we've got 24 equals two X minus two square root of X minus six. Great. Now what can we do to simplify? Well all three of those terms here are divisible by two. So let's take everything and divide it by two divided by two is 12, two, X divided by two is X. And two squared of x minus six, divided by two is square root of x minus six. Okay, that's a lot more simplified. Now we want to get that radical on one side by itself so that we are going to be able to get rid of the radical and because it's negative, I want to move the radical over to the other side. So I'm going to add it to both sides and I want to subtract this 12 to move it over to the right hand side so that the radical squared of X -6 will be alone on the left And then we'll have X -12 on the right hand side. Now we've got the radical by itself and we will square both sides in order to get rid of the radical on the left. So the square root of x minus six squared is just x minus six equals. And now we've got x minus 12 times x minus 12 and we are going to foil that. And I want to pause for a second and make a note that when we took this square root of x minus six and squared it, we now gave ourselves a power of two. We're going to have an X squared and our original problem only had a power of one. The degree was one for the highest degree. It was only two X to the first. So we're going to now come up with two answers whereas before there's only one answer for X. So we're going to get two answers now and one of them we might be able to throw away. We'll have to check both answers when we're finished. So let's go ahead and foil. Now we'll bring down X minus six and so take our first X times X is X squared are outside X times negative is negative 12 X. Our insides negative 12 times X is again negative 12 X. And our last negative 12 times negative 12 is plus 144 will combine those like terms in the middle negative 12 X minus another 12 X will give us x minus six equals X squared minus 24 X plus 144. But we're not done yet. Okay, let's get one side equal to zero so that hopefully we can factor. So let's subtract X from both sides and add six to both sides. So the left hand side is now zero equals X squared negative 24 X minus X. Is negative 25 X. And 144 plus six is plus 150. Okay, we can factor this? We have what times what gives us X squared. That's X times X. What times what gives us 100 and 50 positive 150 but adds up to a negative 25. Well that would be a minus 10 and a minus 15 If you want. You are welcome to go ahead and pause the video and foil that out to double check. But I'm going to continue and I'm going to set both of those factors that we found X -10 and X minus 15 equal to zero. So that we can solve for R two X values for x minus 10 equals zero will add tenable sides. So now we have X equals 10 as one of our solutions and for x minus 15 will add 15 to both sides and that will give us X equals 15 for one of our solutions. But as I mentioned before, we artificially created two X solutions when we've only got a degree of one. So now we have to plug both of those back into the original equation to see if they work. So we've got why? Well we'll put in 24 for y. So 24 equals two X minus the square root of four, X minus 24. And now fill in tens for all of our exes, 24 equals two times 10 minus the square root of four times 10. Oops, that doesn't even look like a 10, I'm not sure what that was Four times 10 minus K 24 equals two times 10 is minus the square root of four times 10 is 40 minus 24. So 24 equals 20 minus the square root of 40 minus 24 that's 16. So the squared of 16, 24 equals 20 minus the square root of 16 is four. And we're getting 24 equals 20 minus four. Or in other words 24 equals 16 and no it does not, 24 does not equal 16. Therefore X cannot equal 10. That one doesn't work. So now let's go back in and plug 15 in to make sure that one works. 24 equals two X minus the square root of four. X minus 24. Or in other words 24 equals two times minus the square root of four times 15 minus 24. So 24 equals two times 15 is 30 minus the square root of four times 15 is 60 minus 24. 24 equals 30 minus well 60 minus 24 is 36. So the squirt of 36 equals 30 minus the square root of 36 is six. 24 equals 30 minus six is 24. Yes, this one does work. So X does equal 15 which gives us our answer choice of B. X equals 15. Well done, we'll catch you on the next one

In Exercises 91–100, find all values of x satisfying the given conditions. $y = x - \sqrt{x - 2} \quad \text{and} \quad y = 4$

Key Concepts

Solving Equations Involving Square Roots

Domain Restrictions for Square Root Functions

Checking Solutions for Extraneous Roots

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