The Square Root Property - Online Tutor, Practice Problems & Exam Prep

Quadratic equations can be solved using various methods, including factoring, the square root property, and recognizing when complex roots arise. For example, the equation \(x^2 + x - 6 = 0\) factors to \((x + 3)(x - 2) = 0\), yielding solutions \(x = 1\) and \(x = -3\). In contrast, \(4x^2 + 25 = 0\) leads to complex solutions \(x = \pm \frac{5i}{2}\) when simplified. Understanding these methods enhances problem-solving skills and prepares students for diverse mathematical challenges.

Solving Quadratic Equations by the Square Root Property

Hey everyone. So we just learned how to solve quadratic equations by factoring. So if I'm given something like x2+x-6=0 then I can just factor that into (x+3)(x-2) and get my solutions from there. And that's because this equation is easily factorable. But if I'm given something like x2-5=0, well, I'm not really sure what the factors would be there and that's actually because this equation is not factorable at all. Not all quadratic equations are going to be able to be solved by factoring, but that's okay because there are actually 3 other methods that we can use to solve quadratic equations. Now I know that that might sound a little intimidating right now and, like, a lot, but I'm going to take you through each of these methods step by step, and soon you'll be able to solve any quadratic equation that gets thrown at you. So let's go ahead and jump in.

So, we're going to start filling out this big table of all of the information that you will ever need to know about quadratic equations. And we already looked at our steps to solve a quadratic equation by factoring, but we need to look at another piece of information here which is when we should use each method because there are multiple. So with factoring, if a quadratic equation had obvious factors or if it has no constant term, so in my standard form equation, if c=0, then factoring is going to be a good choice to solve it. But if these two things are not true, then we're going to need a new method. So here we're going to look at the square root property.

Now if those two things are not true but I'm given something in the form of (x+anumber)2=aconstant, so something like (x+1)2=4. Or if I have no x term, so in my standard form equation b=0 and I have something like 4x2-5=0, then I'm gonna go ahead and use the square root property. Now the square root property tells us that we can solve by literally just taking the square root of our quadratic equation. So let's go ahead and see that in action.

So looking at my first example here, I have (x+1)2=4. So looking at our steps, step number 1 is going to be to isolate our squared expression. So whatever is being squared, I want that by itself. Now here I have x+1 being squared, so I want that by itself and it already is. My squared term is by itself on that left side, so step 1 is done. Now step 2 is going to be to take both the positive and the negative square root. This is what's gonna give us both of our solutions. So if I go ahead and square root my entire quadratic equation, this left side is going to cancel the square, leaving me with x+1. But then on that right side, I need to take both the positive and the negative square root. So this is going to become x+1=±4. Okay. So we can actually simplify this a little bit more. x+1 is equal to ±2. Okay so step 2 is done, now we can go ahead and just solve for x for step 3. So here if I move my one over to the other side by just subtracting it, I'm left with x=±2-1. Now that looks a little funky the way it's written now, so if I just rearrange this and move my one over to the front, I will be left with x=-1±2. Now I'm actually gonna be able to further simplify this if I split it into my plus and my minus and my answer, so let's go ahead and do that. So this really can become -1+2 and -1-2. Those are my two solutions. Now -1+2 is just 1, and then -1-2 is -3. So these are actually my 2 solutions here, which I can rewrite a little bit nicer as x=1 and x=-3. Those are my final solutions, which I solve by just taking the square root.

Let's take a look at one more example. So over here, I have 4x2-5=0. Let's go ahead and start back at step 1, which is to isolate our squared expression. Now here I just have x2, so I want to get that by itself. Now the first thing I need to do is move my 5 over to the other side, which I can do by adding 5. It will cancel, leaving me with 4x2=5. Now one more step to isolate that squared expression is to divide by 4, leaving me with x2=54. Okay. So step 1 is done. We've isolated our squared expression. Now we can go ahead and move to step 2, which is to take our positive and negative square root. So if I go ahead and square root my quadratic equation, it will cancel on this side, leaving me with x=±54. Make sure you don't forget that you're taking both the positive and negative square root here. Okay, so we can actually simplify this a little bit further and ±52. So step 2 is done. We've taken our positive and negative square root, and now we're left to solve for x. But x is actually already by itself here, so I'm done and these are my solutions. Now remember that you can always take your solutions and plug them back into your original equation to check, and I will leave that up to you. But here we're done. Our solution is just ±52. I could always split that into the positive and negative solutions, but it's not going to further simplify it here, so I'll just leave it at that.

Now you might have noticed that our answers looked a little different here. So on one hand, I have x=1 and x=-3, and over here I have the square root of 5 over 2. And that's actually because our solutions are not always going to be whole numbers. Sometimes they might have fractions or even radicals in them or some combination of both and that's totally fine, your solution doesn't have to be a whole number. So that's all you need to know about the square root property, let's get some practice.

Solve the given quadratic equation using the square root property. $\left(x-\frac12\right)^2-5=0$

• A. $x=\frac12+\sqrt5,x=\frac12-\sqrt5$
• B. $x=\frac52,x=-\frac52$
• C. $x=\frac{\sqrt5}{2},x=-\frac{\sqrt5}{2}$
• D. $x=\frac12,x=-\frac12$

Solve the given quadratic equation using the square root property. $2x^2-16=0$

• A. $x=0,x=-2$
• B. $x=4\sqrt2,x=-4\sqrt2$
• C. $x=4,x=-4$
• D. $x=2\sqrt2,x=-2\sqrt2$

Imaginary Roots with the Square Root Property

Hey, everyone. As you solve quadratic equations using the square root property, you may end up getting an imaginary or a complex root. But that's totally fine because we know how to deal with complex numbers, and we're just going to simplify it as we would any complex number. So let's look at an example. Here we have 4xx2+25=0. And I want to solve this using the square root property. So starting with step 1, of course, I want to isolate my squared expression.

Now my squared expression is just this x2 so I want to go ahead and get that by itself by first moving this 25 over and I can do that by just subtracting 25 from both sides. I'm left with 4xx2=−25 and then divide by 4 to get that x2 by itself. I am left with x2=−254. So step 1 is done. I have isolated my squared expression and now I want to take the positive and negative square root.

So taking the square root of both sides here it will cancel on this side. I'm left with x=±−254. Now I can go ahead and simplify this further and I can separate it out into ±−254 just using my radical rules. Now I have a negative under the square root here but it's totally fine because we've seen this before and we know how to deal with it. So let's go ahead and simplify this.

Now this square root of negative 25 is just going to become 5I and then we know that the square root of 4 is just 2. So really, this simplifies all the way into ±5I2. So we have completed step 2. We've taken our positive and negative square root, and x is actually already by itself so step 3 is done as well. And now I just have my solution. So my solution is x=±5I2. And I can, of course, separate these into the positive and negative if I want as well.

Now whenever you're dealing with a complex answer, it's totally fine. It's fine to have the imaginary unit. And a clue that you might end up with a complex answer is if in your standard form equation a and c have the same sign, then you're always going to end up with a complex answer. That's all there is for imaginary roots. I'll see you in the next video.