In this lesson, we're going to extend our knowledge of solving radical equations. This time our radical equations will lead to a quadratic equation. We'll apply the same strategy of isolating the radical on one side, and then we'll raise both sides of the equation to the index of that radical. For example, if we have this radical equation? What will we get if we get if we square the left side and the right side?
When we square both sides of the equation, we get 2x plus 6 on the left and x squared minus 2x plus 1 on the right. Great work if you found these two parts. We know a square undoes the square root, so we would be left with this expression on the left. For the right-hand side, we want to square x minus 1. This means we'll multiply x minus 1 by itself. When we do that, we will get the polynomial x squared minus 2x plus 1. Remember this is really a perfect square so it follows the pattern a squared minus 2ab plus b squared. We can use that pattern to multiply this more quickly.
Now, we're almost to an equation that can be factored. We just need to rearrange the terms, so all of the terms are on one side of the equation. If all the terms are on one side of the equation, that means the other side must equal 0. What I want you to do is to rearrange the equation so that happens. What do you think we'll get here?
If we subtract 2x from both sides and 6 from both sides, then we'll have 0 on the left, and x squared minus 4x minus 5 on the right. Great work if you found this expression. Now, this might look a little funny to you, but I just combined the subtraction in one step. I subtracted 2x and I subtracted 6. Keep in mind that we can do this so long as we perform the same operation on the left as we do on the right.
So, now this is something we should recognize. We have a quadratic equation on the right side, and 0 on the left side. So let's see if you can solve this quadratic equation. When you think you have the answers, enter them here and write your answer as a set.
The solutions to this quadratic equation are negative 1 and positive 5. Nice work if you got these two answers. When we factor this quadratic, we know the first terms in each binomial must be x, since x times x equals x squared. Now we want to find factors of negative 5 that sum to negative 4. Those factors are negative 5 and positive 1. Next, we set each factor equal to 0 and then we solve for x. So, x can equal positive 5, or x can equal negative 1.
Now, we might think we have the solutions here to our original equation, but we always want to check. When squaring an equation or raising an equation to an even power, we need to be careful. We don't want to pick up the negative root, so we must check our answers in the original equation. So, what I want you to do is to check negative one and positive 5 in the original equation. What would you get on the left hand side? And what would you get on the right hand side here? Do the same for this side, and then we'll see the result.
When we check negative 1, we'll get 2 on the left and negative 2 on the right. This is not a true statement, so we know that x equals negative 1 is not going to be part of our final solution. For x equals 5, we'll get 4 on the left and positive 4 on the right, so, yes, this answer does check and will stay as part of our final solution set.
So when solving radical equations, sometimes we'll get a quadratic with two potential solutions. But we need to check these solutions in order to determine which ones actually satisfy the original equation. In this case, negative 1 did not check, so our only solution for this original radical equation is 5. This is our solution set.
Try solving this radical equation. Keep in mind you want to isolate this radical before you square both sides. Once you find two solutions, be sure to check your answers and then write your answer here as a set.
It turns out there's only one value that makes this equation true. And it's positive 12. Great algebra skills for getting this one correct. Now if you didn't get this solution, that's okay. Math can be pretty tough. So, let's figure out how to do this together. First, we want to subtract 9 from both sides, to isolate this radical. Next we'll square both sides to undo this radical, or square root. So, I have x squared minus 18 x plus 81 on the left, and x minus 3 on the right. We subtract x from both sides and add 3 to both sides to get 0 on the right and this expression on the left. Now we want to find factors of 84 that add to negative 19. This factor pair would be negative got stuck. The easiest way to find the factor pairs of 84 is to start with 1 and 84. Then you can just go up to the next number to see if it divides in evenly into 84. I know 84 is even so 2 does go into it. We can try each of the numbers going up and then we can see that 7 and 12 would be the factor pair that we need. They add to positive 19 so they must both be negative if they are going to add to negative 19 and multiply to positive 84... This is how we know this must be our factor pair. Next, we can set each factor equal to 0 and solve for x. So, x could equal 12 or x could equal positive 7. Now, keep in mind that we need to check these solutions. We took the square of both sides, so we might've picked up a root that won't work in the original radical equation. When we check x equals 12 we get 12 equal to 12 so yes, this answer does check. But on the right side with x equals 7 we don't get an answer that checks. We'll have 7 equal to 11 which we know isn't true. So, we remove 7 from our solution set leaving us with just positive 12.
What about this radical equation? What do you think the solutions would be here?
The only solution for this equation, is positive 8. Great algebra skills for getting this one correct. We'll take the same steps to try and solve this. First we'll isolate the radical on one side of the equation. And lucky for us, it already is. Next we'll square both sides, and then try and factor our quadratic. When we square both sides, we'll get ax squared minus 6a plus 9 on the left, and 3a plus 1 on the right. We move these terms to the other side of the equation, so that way we can set it equal to 0. We factor a squared minus negative 8. We solve for the factors of a by setting each factor equal to 0. So, a could be positive 8 or a could be positive 1. Next we want to check positive 8 and positive 1 in the original equation. Plugging in x equals 8, we get a true statement of 5 equals 5. When we plug in x equals 1, we don't end up with a true statement. We'll have negative 2 equals 2. Since this doesn't check, we know x equals 1 is not part of our solution set. We'll just have the answer x equals 8.
Here's our third problem on solving radical equations. What values of x do you think make this equation true? Now, write your answer as a set here, and type ns for no solution, as usual. Good luck.
Here the only value that makes those statements true is positive one. That's amazing work if you got this one correct. It was tough. First, we'll square both sides of the equation to get 4x squared plus 4x plus 1 equaled to 5x plus side equal to zero. So we can subtract 5x from both sides, and subtract 4 from both sides. We'll have 0 on the right and this quadratic on the left. Next, we use factoring by grouping to get 4x plus 3 times x minus 1 equals 0. Once we have our factors of 4x plus 3 and x minus 1, we set each of those equal to 0. Then, we can solve for x, so x equals negative 3 4ths. Or x equals positive 1. And like all of our radical equations from before, we need to check each of these solutions in the original equation. When I plug in x equals 1, I get 3 equals 3. So, yes, this will be part of our solution set. For x equals negative numerator and denominator. Here we'll just multiply 5 times negative 3 4ths to get negative 15 4ths. We change 4 into 16 4ths so we can add these two fractions together. This gives us a positive 1 4th underneath our radical. W stake the radical of the numerator and the denominator to get 1 half since the square root of 1 is 1 and the square root of 4 is 2. On the left side I changed tho simperer fraction to a mice dumber. Negative 1 and 1 half. We know negative statements are not true, so we cannot include this as part of our solution set, it's out. This is why our solution set is only positive 1.
Now we're on to our fourth practice problem. What do you think the solutions would be for this radical equation? Notice we have a constant in front of this radical sign. We can leave this in front of the radical, we just want to make sure when we square both sides, we square the 4 and we square the radical. Try it out on your own, and then see if you can find the correct answers, or maybe there's just one answer, or no answer. Good luck.
Here the solution to our equation is 2. Great work if you got that one orrect. Now if you didn't quite know how to handle this four let's see how to do it, then you can pause the video once you figure out and then continue solving. To start we'll square both sides of the equation so we can undo the square root. We square 4 which equals 16, and then we square the radical which gives us the x minus 1. Just this radicant. On the right side we square the binomial 8 minus the equation to get 16x minus 16. On the right-hand side, I'll just re-arrange the position of these two terms so I have my x squared term first. We'll have Next we want to set our equation equal to 0, so we subtract 16x from both sides. And we add 16 to both sides. We do this since we know we'll get 0 on the left, and this polynomial on the right. We'll have a quadratic that we can try and factor. All of these terms have a common factor of four, so we can divide out our equation by that factor. We'll have 0 equals x squared minus 12x plus are negative 10, and negative two. We set each factor equal to 0, so x can equal 10 or x can equal 2. And finally, we check x equals 10, and x equals 2, and the original equation. When we do this, we see that 10 does not check, we get 12 equal to negative 12. So this one is not going to be part of our solution set. However, for 2, it does check, so this must be our only solution. So don't worry if you stumbled at the beginning with squaring this 4, if you are able to finish it out, great work.
Here's our last practice problem for the lesson. What do you think would be the solution for this radical equation? Write your answer as a set, or if there's no solution, type NS as usual. And, just like our last problem, we have a number in front of our radical. So be sure you square this number when you solve.
Here our solutions set is negative 2 and positive 6. That's amazing work if you've found those two answers. First we want to square both sides of our equation. 2 squared equals 4, times the quantity 2x plus 4, since the square of this square root is simply what's inside here. On the right side, we'll square the binomial x plus 2 to get this expression. Next we'll distribute the 4 to get 8x plus 16 on the left, and our same expression on the right. We want to set our quadratic equal to 0, so we subtract 8x from both sides, and we subtract 16 from both sides. This leaves us with 0 equals x squared minus 4x minus 12. Now, we're ready to try and factor. We find factors of negative 12 that sum to negative 4. Those factors are negative 6, and positive 2. We set each factor equal to 0 to get x equals 6, or x equals negative 2 as our potential solutions to the original radical equation. Finally, we check each of the solutions in the original equation and discover they both check. This means we can include both of these numbers in our answer to have our final solution set.