Solve each radical equation in Exercises 11–30. Check all propose...

Question

Solve each radical equation in Exercises 11–30. Check all proposed solutions.

√(2x + 3) + √(x - 2) = 2

Accepted Answer

Welcome back. I'm so glad you're here. We've got this equation and this one is involved. So kudos to you for trying it. So now let's see, we've got the square root of X plus one. On the left hand side of the equation equals one minus the square root of two X minus ones. We've got square roots on both sides. Now let's work on eliminating those square roots. And so we're going to square both sides and on the left now instead of the square root of X plus one. Well, when we square that, we get rid of the square root, so it's just X plus one equals and now we've got to take what we have on the right hand side and foil that out. So we've got one minus the square root of two, X minus one times one minus the square root of two X minus one. And now we foil that with our first one times one. Well that's one our outsides one times negative squared of two, X minus one is going to be a negative square root of two, X minus one. And the insides. That's the same thing. The negative square root of two x minus one times one, so minus another squared of two, X minus one. And when we take these lasts we have the square root of two x minus one times the squared of two x minus one. Well they're both negative. Now it's a positive square root of parentheses, two X minus one squared. So we bring down the left hand side of the equation. That's still X plus one. We can do some simplification. The square root of something squared. Well that's just going to be the number inside two x minus one. So the squared and the square root cancel out. And these two terms here are like terms, so they add up together and so we have two of them. So we say minus two square roots of two X minus ones and we can bring down that one and bring down the X plus one. Sorry, we just shifted a little to the right And now let's combine like terms on the right, we have a 1 -1. So we have X plus one equals while the one minus one cancels out. So negative two square root two X minus one plus two. X. Let's get that radical by itself on the right hand side so we can get rid of it. So we're going to subtract two X from both sides. And on the left x minus two, X is a negative X plus one equals negative two square root of two, X minus one. Now we'll divide both sides by negative two, divide each term by negative two. And so on the left we have negative X divided by negative two. Well that's going to be X over two plus Well one divided by negative two. That's negative one half equals on the right hand side. The square root of two, X minus one. Can you see why this is I mean this is an involved problem. To get rid of that square root on the right hand side we are going to square it. And what we do to one side we have to do to the other. So on the right hand side it's very nice. It's two X minus one. And on the left. Well now we get to take parentheses X divided by two minus one half times X divided by two minus one half. Fun stuff. It's like do you want to practice every single skill you've ever learned in math? So for our firsts when we foil this, X over two times X over two is X squared over four for our Outsides, X over two times a negative one half will give us minus X over four for our inside same thing negative one half times X over two minus x over four. And for our last negative one half times a negative one half gives us a positive 1/4 equals two x minus one. Well at least there's no more radicals. Huh? And now let's get rid of our fractions. Let's multiply every single term times four. All four of those terms on the left, their denominators will cancel out when we multiply the numerator is by four. So that's just going to be X squared minus x minus x plus one equals two X times four is eight X minus one, times four is four. We can combine these like terms could have done that before. Doesn't matter for the order here we've got x squared minus two X plus one equals eight x minus four. Hey this is looking like a real problem now and let's bring over our eight X to the left and our four. So we'll subtract X from both sides and add four to both sides. And so we've got X squared minus 10 X plus five equals zero. And oh man We can't factor that. There is nothing. We're going to multiply factors of five that will get to add up to negative 10. So now truly this is the problem to review all problems. We're going to pull out our friend that we recall from previous lessons, the quadratic formula. So we are going to have negative B plus or minus the square root of B squared minus for A C. All over two. A. Let's identify our A B and c. A is in front of the X squared term. So A equals one. B is in front of our X terms. So B equals negative 10 and c is our constant term. So C equals five. Let's fill these in. Ok, negative B That's a negative negative 10. So negative negative 10. You can write that as positive 10. I'll do that on the next step plus or minus the square root of B squared. That's negative 10 squared minus four times one times five. All over two times one. Okay, we'll do some simplification, negative or negative negative 10 is plus or minus inside this radical radical again. Okay negative 10 square that's 100 minus four times one times five is 20 Over two times 1 is two 10 plus or minus. The square root of 100 minus 20 is 80 all over to. Okay let's figure 80 out factor tree. If we have 80 that is the same as let's do 10 times 8. 10 is five times two and eight. Well that's two times four. Well there's a two squared so five times two squared for those two twos and four. That's also two squared. So two squared times two squared. Well that's also four squared. So we've got five times four squared the four gets to come out. So that'll be four square root five. So we've got 10 plus or minus four square root of 5/2. We're not done. We can take 10 and four square to five and divide both by two. So that gives us five plus or minus two square root of five. Oh my goodness we just found our two X. Values. Right well actually we don't know I'm going to rewrite this up here so that when I scroll up we can still see it so five plus or minus two squared of five. And so when we did this problem let's look at what we had for our degree to start off with and our degree in the original is a one. We had the highest. Actually it's a one half, Everything is inside of a square root. So those are all X to the one half. And so we artificially created and X squared when we were solving for it way down here. Actually we got an X squared at this point, which would give us two answers when we're solving for X. But because we artificially created those, we might be creating extraneous solutions, meaning ones that aren't correct that don't apply. So it's possible that X equals five plus two squared to five or five minus two square to five. But because we created that X squared in our process, we have to go through and check and I am going to scroll back down where we've got more room but now we've got that answer at the top and we can see our answer choices. So scrolling back down, oh no. You know what? I'm going to rewrite our problem so that I can bring that down. So we've got the square root of X plus one equals one minus the square root of two, X minus one. So this is the color we're using to check the solutions. Let me see if I can capture that without capturing anything else and bring that down where we've got room. Okay, so that is our original and we're going to fill in our five plus two squared to five for an X. And then our five minus two squared to five for X. Seriously this problem is amazing. It's it's I'm in awe of it so it's the square root of we'll start with the positive. So five plus two squared to five plus one equals one minus the square root of two times five plus two square to five minus one. Oh and end that parentheses right there because that's what the two is being distributed to. Okay if you are adventurous and curious you can go through and solve that. Um Just doing the algebra and distributing the two and getting rid of our radicals like we did earlier. It's very similar process to what we went through in the black coloring. But if you have this available to you I highly recommend using a calculator at this point and you can plug this in the square root of five plus two square to five plus one and that's going to give you 1.23607. And just I mean spoiler alert pause here if you want to go through and do the whole math algebra yourself so that you don't know the answer. But if you're not going to pause then we plug this other side into a calculator, one minus squared of two parentheses five plus two square to five. End parentheses minus one. And oh I copied the wrong thing. I had these all written down, I did not memorize them. The left hand side is 3.23607. I apologize. The right hand side of the equation is negative 3.23607. That means x does not equal five plus two square to five. That was an extraneous solution that we created in doing all of this work. And now let's go ahead and try 5 -2 square to five. That one might work. Give ourselves a little more room again. And so we've got the square root of yes five minus two square to five plus one equals one minus the square root of two times five minus two squared to five. And parentheses -1. And we put that in a calculator and the left hand side of the equation, that's where we get this 1. equals. And the right hand side of the equation. Drum roll, it is 0.76393. And they go on after that I rounded all of these decimal points but yeah X does not equal 5 - square to five. That was also an extraneous solution that we created as we were squaring all of those radicals. So we scroll back up to the top and we did all of this work. And if you went through an algebraic lee solved for those last two checking the solutions, you did a lot of work all for d There is no solution. Well done. We'll catch you on the next one.

Solve each radical equation in Exercises 11–30. Check all proposed solutions. √(2x + 3) + √(x - 2) = 2

Key Concepts

Radical Equations

Isolating the Radical

Extraneous Solutions

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