We're going to use our knowledge of radicals and factoring to solve quadratic equations. We'll use three different strategies in this unit, and we'll learn them one by one. The three methods are the square root method, completing the square and the quadratic formula. First, we'll learn each of the methods, and then you'll have a chance to pick which method you think is best to use in different cases. For this lesson let's start by focusing on the square root method.
If a quadratic equation is in this form, we can simply divide both sides by a, and then take the square root of both sides to find x. If we divide both sides by a, we'll get x squared is equal to c divided by a. And now we take the square root of both sides. When we take the square root of both sides, we need to add a plus or minus symbol in front of the side that doesn't have a variable. And here's why we include this plus or minus sign. Think about if I had x squared equals 1. I know there are really two values that make this true. Positive 1 squared would equal 1, and negative 1 squared would also equal 1. This means when we solve and take the square root of both sides. We need to add the plus or minus symbol to account for both the positive and the negative answer. The square root of x squared is x and the square root of 1 is 1. So really x could be positive 1 like we thought or x could equal negative 1. So you see if you can use the square root method. Try solving this equation by following these steps.
First, we divide both sides of our equation by 5, to get x squared equals positive 6. Next, we take the square root of both sides and we indicate the plus or minus sign here. This gives us our answer of positive root 6 and negative root 6. You could have also written the answer like this, root 6, negative root 6. Or switch the order of these 2. So solving this equation isn't that much harder, you just have new symbol of plus or minus.
Like all of our other equations, we can check to make sure that this solution is correct. If we substitute in positive root 6 into our equation, we'll notice that it checks. The square root of 6 squared equals 6, and 5 times 6 equals 30. But remember, we can have the positive answer or the negative answer. So let's check x equal to negative root 6. We plug in negative root 6 in for x, and square it. Squaring a negative number always results in a positive. So we know negative root 6 squared equals positive 6. Then we have 5 times 6, which is 30, so our answer checks. This is always a great way to check your answers. ANd this reinforces why we have the plus or minus symbol. It's because we have two answers.
Since we know that taking an even root of an equation creates a positive and a negative result, we need to be able to enter this new sign, this plus or minus. Luckily Math Quill has a way for us to do that. To enter this plus or minus sign, you want to type in a backslash, and then the letters p and m. Of course, p and m here stands for plus or minus. But like most things in math quill, we also need to enter a space in order for this symbol to appear. Try typing these first two expressions using the keystrokes on the screen. And then see if you can figure out these last two. Keep learning, and good luck.
For the second one, you need to type in the nthroot and a space. This will create this radical symbol. Your cursor bar will appear here, and you'll need to hit the Right Arrow key and then type in 5 to get what's underneath the radical. These would be the keystrokes for our third expression. Notice that after we type in the square root symbol, we don't need to hit the Right Arrow key. That's because the index for this root is assumed to be a 2, It's rare that you'll ever see it, but we know it's a square root. That's why it's a 2. Math Quill knows it's 2, and so you're just placed back inside the radical. And then you can type in 21. Now for this last expression, you might not have any idea what i is. That's okay. We're going to cover this later. To start this expression, I would type in a forward slash. This would create an entire fraction with a spot for a numerator and the denominator. We type in these keystrokes in order to get our entire numerator. And notice that our cursor bar is here. Once we type in the 6, our cursor bar will appear here, and then we hit the Right Arrow key twice in order to move it down to the denominator. Then, we just type a 4 to wrap up our expression. Now, there are other ways that you can have done this to recreate the fraction bar. You could've created the fraction bar after the 6. But that means, this forward slash would go in between these 2 errors. After you type the 6, you're still inside the radical. So, if you type the forward slash here, you'll create a fraction within the radical. You don't want that to happen, so you hit the Right Arrow key once. Which moves the cursor bar outside of the radical, and then you can type a forward slash to create your fraction, hit the Right Arrow key again to move to the denominator and then enter 4 to finish up. Either way you do this last one is correct, I just want you to get comfortable using math quill.
If that first one gave you some trouble, try this one out. What do you think the solutions are for x? Again, you can either enter your answer with a plus or minus sign, or you can write your two answers separated by a comma.
Here, the answer is plus or minus 5 times root 2. Great thinking if you found these answers. We start by dividing both sides of our equation by 4. This gives us x squared on the left, and 50 on the right. Whenever we use the square root method, we want to isolate x squared. Or there might just be some sort of quantity squared. This allows us to take the square root in this step. Doing that we get x is equal to positive or negative square root of 50. Then we just use our knowledge of simplifying roots to get x is equal to plus or minus 5 times the square root of 2.
Our square root method also works for equations in this form. Notice that we just have a quantity over here, squared equal to some number. This means we can take the square root of the left side and the right side. When we take the square root of this side, it undoes the square power, so we're just left with this quantity, ax plus b. So, I have ax plus b on the left side of our equation and plus or minus root c on the right side of our equation. Next we subtract b from both sides, so that way we can start to isolate the X. Notice that this negative b, is not like terms with radical c. All I can do, is list this as negative b plus or minus the square root of c. Finally, to get x by itself, we divide both sides by the coefficient of a. So we know x could equal negative b plus root c divided by a. Or x could equal negative b minus root c, all divided by a. There's really two solutions here.
So, using this as your guide, what do you think the solutions would be for this equation? You want to write your answer here using the plus or minus symbol, or you can enter them separated by a comma. Now not every single step on this left side will match this side, but I think you can figure the steps out. Good luck.
Our solutions for x are 5 plus or minus 2 times root 3. Great thinking if you got this. Now, I know we haven't done that much practice yet. So that's totally okay if you struggled. I'm sure you'll get the next one. We begin to solve this problem by taking the square root of both sides. Since we have a quantity squared, with a variable inside it, on the left. Taking this square root undoes the square power. Leaving us with x minus 5 on the left hand side of our equation. On the right hand side, we'll just have plus or minus the square root of 12. Next, we'll add 5 to both sides to isolate our x variable. So, we'll have x is equal to 5 plus or minus the square root of 12. And again, here this list them as a sum and difference. We need to simplify the root of 12, so we break it into 4 times 3, this gives us our final answer of 5 plus or minus 2 root 3.
All right. Let's see if you have the right stuff. Try solving this equation and remember to simplify your answer in the end if possible.
The solutions are negative 2 plus 3 root 2, and negative 2 minus 3 root 2. Great thinking if you found those two answers. We start by taking the square root of both sides of our equation. The square root of the quantity x plus 2 squared is just x plus 2. And when we take the square root of 18, we pick up the negative root. So we have plus or minus the square root of 18. Next we isolate x, or get it alone by subtracting 2 from both sides. We don't have light terms here, so we have x equal to negative 2 plus or minus the root of square root of 9 to get our final answer. And do keep in mind there's really two answers here: we have negative 2 plus 3 root 2 and negative 2 minus 3 root
How about this one? What do you think the solutions would be for x? Now there's actually one more step at the end here. Maybe you can get it. As a hint, we no longer have coefficient of 1, in front of x. Now it's a 3.
Here the solutions are negative 5 plus or minus 2 root 6 all divided by 3. Great algebra skills if you got this one correct. You may have also entered this as your answer. This is correct as well. You would of divided 3 into the first term and 3 in the second term. Since there's no common factors we just list these two terms as fractions. With a plus or minus symbol in between. To start this problem, we take the square root of both sides of our equation. And we indicate the plus or minus sign for this root. So we'll have 3x plus 5 on the left. And positive or negative root 24 on the right. Next, we want to isolate this 3x by subtracting 5 from both sides. We'll get 3x equals negative the coefficient 3. This leaves us with 1x equal to negative 5 plus or minus root 24, all divided by 3. We want to look to see if we can simplify any radicals, if possible. The square root of 24 is simplifies to 2 times root 6. So we can replace this here in our final answer.
Sometimes we'll need to perform some extra steps to isolate the square before taking the square root, like in this equation. We have something next to our square. We want to get this square quantity alone, so that way we can take the square root of both sides of our equation. So what do you think we'll need to do here? When you think you have the solutions for x ,enter them here using your plus or minus sign. And don't forget to simplify your root if you can. Good luck.
The solutions for x are 3 plus or minus 4 root 3, all divided by 2 great algebra work for getting that one correct. First we want to isolate the square by adding 48 to both sides of our equation. So, we'll keep this squared quantity on the left and we'll have positive 48 on the right. Next we take the square root of both sides and indicate the plus or minus root here. This square root undoes our square so we're left with the quantity 2x minus three. On the right we still just have plus or minus root 48. We add three to both sides to isolate our 2x, and then we divide both sides by 2, this coefficient. When we divide both sides by 2 we're left with 1x on the left and 3 plus or minus root get this answer. Keep in mind you could divide 2 into each of these terms. This would give us x equal to 3 halves plus or minus 2 root 3. If you list your answers like this, or if you list your answers like this, they're both correct.
If the last problem was challenging, then see if you can get this one. Isolate this square first and then continue to solve as usual. When you think you've got it, enter your answer here.
The solutions here are 3 plus or minus root 10, all divided by 2. Great work, using that square root method, if you got this one correct. We start solving this equation by adding 10 to both sides. This leaves us with the square isolated on the left and positive 10 on the right. Now that we have something squared alone, we can take the square root of both sides. This means we will have 2x minus 3 on the left. And plus or minus root 10 on the right. We add 3 to both sides to get 2x by itself and then divide both sides by positive 2. This leaves us with 1 x in our answer. And again if you really wanted to, you could split this up into two separate fractions, 3 halves plus or minus root 10 divided by 2. Either form is the same.
Try solving this first practice problem. What are the possible values for X here? You can either enter the value separated by a comma, or you can use our fancy plus or minus symbol.
Here our solution is plus or minus 4. We start by taking the square root of both sides of our equation. When we take the square root, we need to indicate plus or minus, since we could have the positive value or the negative value. We take the square root of 16 to get positive 4 or negative 4. These should also make sense because if we square 4, we get 16. And if we square negative 4, we get 16. Great thinking for finding those answers.
What about this equation, what do you think the values of m could be? For this one you should have two seperate answers. So write one here and the other here, separated by a comma.
The square quantity is isolated on one side of our equation. So, we can take the square root of this side and this side. The square root undoes the square, leaving us with m plus 2 on the left, and positive or negative root 36 on the right. We know the square root of 36 is just 6. So we have this equation next. We subtract 2 from both sides, to get m is equal to negative 2, plus or minus have 2 equations. M could equal negative 2 plus 6, or m could equal negative 2 minus 6. There's really 2 values of m here. And since these are like terms, we can simplify. So m could equal positive 4 or m could equal negative 8. These were the two solutions.
For this third practice problem, try finding the solution for r. You can enter your answer here. For this one, you don't want to enter it, separated by commas. You want to use the plus or minus symbol. Good luck solving this one, I know you can do it.
These are the correct answers. Great algebra solving if you got them. We start by taking the square root of both sides of our equation. So, 3r minus 2 will equal positive or negative root of 27. I'm going to go ahead and simplify the square root of 27 to be 3 root 3. Next we add 2 to both sides of our equation to get 3r is equal to 2 plus or minus 3 root 3. We divide both sides by the coefficient of 3 to get our final result. And if you prefer, you can write your answer as two fractions. 2 3rds plus or minus 3 root 3 divided by 3.
Alright, how about for this last one? What do you think would be the solutions for x, here? Be careful when you solve, and make sure you check your answers.
Here our solution is plus or minus three i an imaginary solution, pretty cool. We take the square root of both sides to get x is equal to plus or minus the square root of negative nine. Now we've simplified this radical and remember that negative square roots, need to be simplified using i. We can break this radical down into the square root of 9 times the square root of negative 1. Then we replace the square root of negative 1 with i. We can also check our result. We know the answers are not 3 and negative 3. Since 3 squared would equal positive 9 not negative 9. Also at negative 3 squared would be positive out, lets make sure the plus or minus 3i works in our equation. I'm going to check our positive 3i here in this equation and our negative 3i here in this equation. When we square the quantity of 3i, we get 3 squared which is 9 and i squared. When we square negative 3i negative 3 times negative 3 is positive 9, and i squared is i squared. And remember back from before, i squared equals negative 1. So we can substitute that in here. This effectively changes the sign of these terms. This gives us a check of negative 9 equal to negative 9 in both cases. So we could be confident knowing that our answers are correct.