Solving quadratics by taking square roots examples | High School Math | Khan Academy

So pause the video and see if you can solve for x here. Figure out which x values will satisfy this equation. All right, let's work through this, and the way I'm going to do this is I'm going to isolate the (x + 3) squared on one side. The best way to do that is to add 4 to both sides.

So adding 4 to both sides, we'll get rid of this 4, this subtracting 4, this negative 4 on the left-hand side, and so we're just left with (x + 3) squared. On the right-hand side, I'm just going to have (0 + 4). So (x + 3) squared is equal to 4.

Now, I could take the square root of both sides. Another way of thinking about it is, if I have something squared equaling 4, I could say that that something needs to either be positive or negative 2. One way of thinking about it is I'm saying that (x + 3) is going to be equal to the plus or minus square root of that 4. Hopefully, this makes intuitive sense for you.

If something squared is equal to 4, that means that this something—right over here—is going to be equal to the positive square root of 4 or the negative square root of 4, or it's going to be equal to positive or negative 2. We could write that (x + 3) could be equal to positive 2, or (x + 3) could be equal to negative 2.

Notice if (x + 3) was positive 2, then (2^2) is equal to 4. If (x + 3) was negative 2, then ((-2)^2) is equal to 4. So either of these would satisfy our equation.

If (x + 3) is equal to 2, we could just subtract 3 from both sides to solve for (x), and we're left with (x) is equal to negative 1. Over here, we could subtract 3 from both sides to solve for (x); so (x) is equal to negative 2 minus 3, which is negative 5.

So those are the two possible solutions, and you can verify that. Take these (x) values, substitute them back in, and then you can see. When you substitute (x) equals negative 1, then (x + 3) is equal to 2. (2^2) is 4 minus 4 is 0.

And when (x) is equal to negative 5, (negative 5 + 3) is negative 2. Squared is positive 4 minus 4 is also equal to 0. So these are the two possible (x) values that satisfy the equation.

Now, let's do another one that's presented to us in a slightly different way. So we are told that (f(x)) is equal to (x - 2) squared minus 9. Then we're asked, at what (x) values does the graph of (y) equals (f(x)) intersect the x-axis?

If I'm just generally talking about some graph, I'm not necessarily going to draw that (y) equals (f(x)). So if I'm just—so that's our y-axis, this is our x-axis—and so if I just have the graph of some function that looks something like that, let's say. That's the (y) is equal to some other function, not necessarily this (f(x)), (y) is equal to (g(x)).

The (x) values where you intercept—the points where you intersect the x-axis. Well, in order to intersect the x-axis, (y) must be equal to 0. So (y) is equal to 0 there. Notice our (y)-coordinate at either of those points are going to be equal to 0, and that means that our function is equal to zero.

So figuring out the (x) values where the graph of (y) equal (f) intersects the x-axis is equivalent to saying, for what (x) values does (f(x)) equal zero? We could just say for what (x) values does this thing right over here equal 0? Let me just write that down.

We could rewrite this as (x - 2) squared minus 9 equals 0. Well, we could add 9 to both sides. So we could get (x - 2) squared is equal to 9. Just like we saw before, that means that (x - 2) is equal to the positive or negative square root of 9.

So we could say (x - 2) is equal to positive 3, or (x - 2) is equal to negative 3. If we add 2 to both sides of this, we get (x) is equal to 5, or (x) is equal to negative 1.

You can verify that if (x) is equal to 5, (5 - 2) is 3. Squared is 9 minus 9 is 0. So the point (5, 0) is going to be on this graph.

Also, if (x) is equal to negative 1, (-1 - 2) is negative 3. Squared is positive 9 minus 9 is 0. So also the point negative 1, 0 is on this graph. So those are the points where, those are the (x) values where the function intersects the x-axis.