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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.

## Rearrange to Solve the Quadratic Equation

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?

## Rearrange to Solve the Quadratic Equation

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.

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.