Equations with Square Roots
Equations with square roots. The test sometimes gives us an equation to solve involving the square root. In such an equation, the variable will appear under the radical. So for example, this would be an equation with square roots, square root of x + 3 equals x- 3. We'll solve this one later in the video.
So this is the type of equation we'll be talking about in this lesson. Certainly, we undo a square root by squaring, and we are always allowed to square both sides. Sometimes with the simplest right of equations, all we have to do is square both sides. So for example, if we had something like square root of x + 2 equals 3.
Well, just square both sides. We get x + 2 on the left, we get 9 on the right, subtract, and we get x = 7, fantastic. But that equation was a bit too simple to appear on the test. The test is not actually gonna hand us something simple on a silver platter like that.
It's gonna be a little trickier. Before we go on, let's think about this. Is it always true for any value of k that if we take the square root of k squared that we'll get back to k? That is to say that the square root undoes the squaring and always takes us right back to where we started from.
Is that always true? And of course, the answer is no. The equation is true for positive numbers and for zero, but not for negative values of k. For example, if k =- 4 then of course when we square it, we'll get positive 16, -4 squared is positive 16.
And when we take a square root of 16 we get 4. In other words, we don't go back to the original starting number. So that's important. This suggests we may run into some kinds of problems when negative values arise. So this is on our radar. What happens when we get negative value?
In particular, when the thing under the radical is negative. We have to pay attention to this. It turns out that in radical equations, we have to be aware of extraneous roots. When we do all of our algebra correctly, including squaring both sides of the equation, the algebra can lead to answers that don't actually work in the original equation.
These are extraneous roots. So I want to emphasize this is not about making a mistake. In other words, even if we do all the algebra correctly, just by virtue of the fact that we square, we produce extra roots, extraneous roots that are not ones that actually solve the original equation. It's important to understand extraneous roots will rise in a radical equation even if you do all the algebra correctly.
Now we can look at the equation we had at the beginning. So here's the equation from the beginning. So of course, what we'll do is we'll square both sides, of course that binomial squared on the right side, we foil that out to x^2- 6x + 9. You may remember the pattern for the square of a difference.
Then we'll gather everything on one side so we get a quadratic equal to zero. Well' factor that, it's very easy to factor and we get two roots, (1, 6). Now, normally with algebra, you'd think, okay, we must be done. We found the value of x. But with rattle equations we have to be careful. Do we know that both of these roots work, maybe they both do.
Or maybe one of them is an extraneous root. So we have to check our answers. We have to check each answer we found to make sure they work because right now just looking at them, (1, 6), we don't know are those both true roots? Are they both extraneous roots? Do they work in the original equation?
The only way we find out is by plugging them in. So here's the original equation. Here are the roots that we found from the algebra. So first of all, we're gonna check the first one, x = 1. Plug it into the left side, we get square root of 1 + 3, square root of 4 which is 2.
Plug it into the right side we get 1- 3, which is negative 2. So the two sides of the equation are not equal. One side equals 2, one side equals -2. So this root doesn't work. Now, we'll check the other one. Plug it into the left side, we get square root of 6 + 3, of course that's 9, square root of 9 is 3.
On the other side we get 6- 3, which is also 3. The two sides work so that one does legitimately work and it does solve the problem. So this equation has one solution that works, x = 6. That is the only solution that works, x = 1 is an extraneous root. Because even though we followed the algebra correctly, and even though the algebra gave us that root, that root does not actually work in the original equation.
We need to square both sides to undo the radical but this very act can produce extraneous roots. If we get a quadratic after squaring, which is common on the test, the algebra will lead to two roots. Sometimes both roots work. Sometimes one root works and one is extraneous.
Sometimes both are extraneous and the equation has no solution. So here's a practice problem. Pause the video and then we'll talk about this. Okay, so here we have a radical on both sides, radical equals radical, so of course we're just gonna square both sides. We get 2x- 2 = x- 4.
Well, a very easy equation to solve, and we get x = -2. All right, very good, but now what happens here if we plug this back into the original equation? When we plug this in, this results in the square root of a negative on both sides. So we get the square root of -6, and square root of -6 is something outside the real number system.
It does not live anywhere on the number line so we can't do math with that. For our purposes that is just an error, and this equation has no solution. Finally, keep in mind that we should square both sides only when the radical is by itself on one side of the equation. If the radical appears with other terms on one side, we'll have to isolate the radical on one side before it would make sense to square both sides.
So here's a practice problem. Pause the video and then we'll talk about this. Okay, so we do not have the radical by itself, so the very first thing we have to do is subtract that two from both sides. So we get the radical 4- 3x = x- 2. Now we can square both sides.
And of course we get the square of the difference, the square of that binomial, and that expands out to x^2- 4x + 4. Now we're gonna subtract 4 from both sides and add 3x to both sides. And this will lead us to x^2- x. Very easy to factor that, that factors to x( x- 1). And the algebra leads us to the solutions x = 0 and x = 1.
Now we need to check these answers. Okay, so those are the roots of the algebra found for us. First of all, check x = 0. Plug this into the left side and what we get is 2 + 4 which is 2 + 2, which is 4. Plug in into the right side, it's 0 and of course 4 does not equals 0, so this one does not work.
So this would be an extraneous route. Now check x = 1, plug this into the left side, we get 2 + 4- 3 * 1, so 4- 3. And of course that would be 1, and so that's gonna be 2 + 1 which is 3. And of course this does not equal 1, does not equal the x which is 1 on the other side of the equation.
So this one doesn't work either. And so neither one of the roots that the algebra gave us works. So this equation simply has no solution. Both the solutions that the algebraic gave were extraneous roots. In summary, to undo a radical in an equation, we need to square both sides. We have to move something else to the other side sometimes to isolate the radical before squaring.
In other words, we need the radical by itself, so if there are other terms on that side with the radical, we need to get rid of them, move them to the other side before we can square. And the very act of squaring produces extraneous roots. Therefore, we must check each answer the algebra gives us back in the original equation.
Read full transcriptSo this is the type of equation we'll be talking about in this lesson. Certainly, we undo a square root by squaring, and we are always allowed to square both sides. Sometimes with the simplest right of equations, all we have to do is square both sides. So for example, if we had something like square root of x + 2 equals 3.
Well, just square both sides. We get x + 2 on the left, we get 9 on the right, subtract, and we get x = 7, fantastic. But that equation was a bit too simple to appear on the test. The test is not actually gonna hand us something simple on a silver platter like that.
It's gonna be a little trickier. Before we go on, let's think about this. Is it always true for any value of k that if we take the square root of k squared that we'll get back to k? That is to say that the square root undoes the squaring and always takes us right back to where we started from.
Is that always true? And of course, the answer is no. The equation is true for positive numbers and for zero, but not for negative values of k. For example, if k =- 4 then of course when we square it, we'll get positive 16, -4 squared is positive 16.
And when we take a square root of 16 we get 4. In other words, we don't go back to the original starting number. So that's important. This suggests we may run into some kinds of problems when negative values arise. So this is on our radar. What happens when we get negative value?
In particular, when the thing under the radical is negative. We have to pay attention to this. It turns out that in radical equations, we have to be aware of extraneous roots. When we do all of our algebra correctly, including squaring both sides of the equation, the algebra can lead to answers that don't actually work in the original equation.
These are extraneous roots. So I want to emphasize this is not about making a mistake. In other words, even if we do all the algebra correctly, just by virtue of the fact that we square, we produce extra roots, extraneous roots that are not ones that actually solve the original equation. It's important to understand extraneous roots will rise in a radical equation even if you do all the algebra correctly.
Now we can look at the equation we had at the beginning. So here's the equation from the beginning. So of course, what we'll do is we'll square both sides, of course that binomial squared on the right side, we foil that out to x^2- 6x + 9. You may remember the pattern for the square of a difference.
Then we'll gather everything on one side so we get a quadratic equal to zero. Well' factor that, it's very easy to factor and we get two roots, (1, 6). Now, normally with algebra, you'd think, okay, we must be done. We found the value of x. But with rattle equations we have to be careful. Do we know that both of these roots work, maybe they both do.
Or maybe one of them is an extraneous root. So we have to check our answers. We have to check each answer we found to make sure they work because right now just looking at them, (1, 6), we don't know are those both true roots? Are they both extraneous roots? Do they work in the original equation?
The only way we find out is by plugging them in. So here's the original equation. Here are the roots that we found from the algebra. So first of all, we're gonna check the first one, x = 1. Plug it into the left side, we get square root of 1 + 3, square root of 4 which is 2.
Plug it into the right side we get 1- 3, which is negative 2. So the two sides of the equation are not equal. One side equals 2, one side equals -2. So this root doesn't work. Now, we'll check the other one. Plug it into the left side, we get square root of 6 + 3, of course that's 9, square root of 9 is 3.
On the other side we get 6- 3, which is also 3. The two sides work so that one does legitimately work and it does solve the problem. So this equation has one solution that works, x = 6. That is the only solution that works, x = 1 is an extraneous root. Because even though we followed the algebra correctly, and even though the algebra gave us that root, that root does not actually work in the original equation.
We need to square both sides to undo the radical but this very act can produce extraneous roots. If we get a quadratic after squaring, which is common on the test, the algebra will lead to two roots. Sometimes both roots work. Sometimes one root works and one is extraneous.
Sometimes both are extraneous and the equation has no solution. So here's a practice problem. Pause the video and then we'll talk about this. Okay, so here we have a radical on both sides, radical equals radical, so of course we're just gonna square both sides. We get 2x- 2 = x- 4.
Well, a very easy equation to solve, and we get x = -2. All right, very good, but now what happens here if we plug this back into the original equation? When we plug this in, this results in the square root of a negative on both sides. So we get the square root of -6, and square root of -6 is something outside the real number system.
It does not live anywhere on the number line so we can't do math with that. For our purposes that is just an error, and this equation has no solution. Finally, keep in mind that we should square both sides only when the radical is by itself on one side of the equation. If the radical appears with other terms on one side, we'll have to isolate the radical on one side before it would make sense to square both sides.
So here's a practice problem. Pause the video and then we'll talk about this. Okay, so we do not have the radical by itself, so the very first thing we have to do is subtract that two from both sides. So we get the radical 4- 3x = x- 2. Now we can square both sides.
And of course we get the square of the difference, the square of that binomial, and that expands out to x^2- 4x + 4. Now we're gonna subtract 4 from both sides and add 3x to both sides. And this will lead us to x^2- x. Very easy to factor that, that factors to x( x- 1). And the algebra leads us to the solutions x = 0 and x = 1.
Now we need to check these answers. Okay, so those are the roots of the algebra found for us. First of all, check x = 0. Plug this into the left side and what we get is 2 + 4 which is 2 + 2, which is 4. Plug in into the right side, it's 0 and of course 4 does not equals 0, so this one does not work.
So this would be an extraneous route. Now check x = 1, plug this into the left side, we get 2 + 4- 3 * 1, so 4- 3. And of course that would be 1, and so that's gonna be 2 + 1 which is 3. And of course this does not equal 1, does not equal the x which is 1 on the other side of the equation.
So this one doesn't work either. And so neither one of the roots that the algebra gave us works. So this equation simply has no solution. Both the solutions that the algebraic gave were extraneous roots. In summary, to undo a radical in an equation, we need to square both sides. We have to move something else to the other side sometimes to isolate the radical before squaring.
In other words, we need the radical by itself, so if there are other terms on that side with the radical, we need to get rid of them, move them to the other side before we can square. And the very act of squaring produces extraneous roots. Therefore, we must check each answer the algebra gives us back in the original equation.
Related Blog Posts
