Completing the Square: Study with Video Lessons, Practice Problems & Examples

To solve quadratic equations like x^2 + 6x = -7, we can complete the square. This involves rewriting the equation in the form x + 62 = c. By adding 62^2 to both sides, we create a perfect square trinomial, allowing us to apply the square root property. This method is effective when the leading coefficient is 1 and the linear coefficient is even, leading to solutions like -3 ± √2.

Solving Quadratic Equations by Completing the Square

Hey everyone. We just learned that whenever we have an equation in the form of x2+bx=c, we can solve it using the square root property and simply square root both sides. But what if I have an equation that I've already determined I can't factor but it's also not quite in the form that I need it to be to use the square root property? Something like x2+6x=−7. Well, I can actually take this equation and force it to be in that form x+anumber2 equals a constant so that I can then use the square root property, something that I already know how to do. But how do we get it into that form? Well, I can actually just go ahead and add some number to both sides to make this x2+6x+something factor to x+anumber2. And this is referred to as completing the square. So I'm going to show you exactly what you need to add to both sides in order to complete the square and then where we go from there. Let's go ahead and get started. So the first thing that we want to look at is when we're actually going to be able to use completing the square and you can actually always complete the square. It's one of these methods that's always going to work on any quadratic equation but there are some instances that it's going to be better to use. So if my leading coefficient a is 1 and b is even, then completing the square is going to be a great choice. So let's go ahead and look at an example of what it actually means to complete the square. So down here I have x2+6x=−7. Now let's check that we would even want to complete the square here. Well my leading coefficient is definitely 1 and then b is 6 here so that is an even number. Completing the square is going to be a great choice. So let's go ahead and take a look at step 1, which is to simplify our equation to this form, x2+bx=c. Now two notable things about this form are that my leading coefficient is 1, and c, my constant, is on the right side of that equation by itself. So let's go ahead and check that we have that form here. So x2 definitely already has a leading coefficient of 1 and my 2, we're going to add b over 2 squared to both sides. Now this might seem really random, why are we adding this b over 2 to both sides? But I'm going to show you exactly how it works. Let's see what happens. So here b is 6 so if I take 6/2 and then square it, that's going to give me 32 which is just 9. So let's go ahead and add that 9 to both sides. So this gives me x2+6x+9=−7+9. So we've completed step 2. We've added that b over 2 to both sides. Step 3 is going to be to factor this to x+b/2 squared. So this x2+6x+9 actually factors perfectly into x+b/2squared2. Here my b was 6 and b/2 is just 3, 6/2, so this factors into x+3squared2 and that's equal to −7+9 which is just 2. So this is always how it's going to work. It is always going to factor perfectly into that b/2 and we've completed step 3. Our final step here is going to be to solve using the square root property. So you might have noticed that our equation is now in the form x+anumber2 equals a constant, meaning that we can just use the square root property as we already know how. So our steps for the square root property are right here. Let's go ahead and start with step 1, which is to isolate our squared expression. Now my squared expression is actually already by itself here, that x+3squared2. So step 1 is done, and I can go ahead and move on to step 2, which is to take my positive and negative square root. So square rooting both sides, I am left with x+3=±2. So step 2 is done as well. I've taken the positive and negative square root. Let's go ahead and solve for x. Now I can do that by moving my 3 over, which if I subtract 3 from both sides, I am then just left with x on my left side there. And then I have ±2-3. Now we can make this look a little bit nicer by moving our −3 to the front there, leaving me with −3 ± 2, and I'm done. Those are my solutions. −3 ± 2 and we've completely finished completing the square. Now let's think about why that works for a second. So when we added this b/2 squared to both sides, what we were really doing is making this into a perfect square trinomial which is something you might remember from our factoring formulas and it's totally okay if you don't. You just need to know that this will always allow this to factor down into x+2 squared, and that's going to be equal to some constant allowing us to just use the square root property. So let's go ahead and complete the square one more time here. And here I have x2+8x+1=0. So let's start back at step 1, which is to simplify our equation to x2+bx=c. Now here I already have that leading coefficient of 1, but let's go ahead and move our constant over to the other side, which we can do by subtracting it. So it will cancel, leaving me with x2+8x=−1. Okay. So now we can move on to step 2 and add b/2 squared to both sides. Now here, b is 8. So if I take 8/2 and then square it, that's going to give me 42, which is just 16. So I'm gonna go ahead and add 16 to both sides. Now when I add 16 to both sides step 2 is completed, and I can go ahead and move on to step 3 which is to factor this into x+b/2 squared. Now we already said that our b/2 was this 8/2 or 4 so, this will factor into x+4squared2 and that's equal to −1 + 16 which is just 15. Okay, so now that we've completed step 3 all we have left to do is to complete solving using the square root property and I'm going to leave that up to you here. So that's all you need to know about completing the square, let's go ahead and get some more practice.

Solve the given quadratic equation by completing the square. $x^2+3x-5=0$

• A. $x=-\frac32,x=\frac52$
• B. $x=-\frac32,x=\sqrt{29}$
• C. $x=\frac{-3+\sqrt{29}}{2},x=\frac{-3-\sqrt{29}}{2}$
• D. $x=\frac{3+\sqrt{29}}{2},x=\frac{3-\sqrt{29}}{2}$

Solve the given quadratic equation by completing the square.

$3x^2-6x-9=0$

• A. $x=3,x=-1$
• B. $x=3,x=1$
• C. $x=2,x=3$
• D. $x=-3,x=-4$

Here’s what students ask on this topic:

What is completing the square in algebra?

Completing the square is a method used to solve quadratic equations by rewriting them in the form ${x+a}^2=c$. This involves adding a specific value to both sides of the equation to create a perfect square trinomial. For example, for the equation $x^2+\; 6x=\; -7$, you add ${\frac{b}{2}}^2$ to both sides, where $b$ is the coefficient of $x$. This method is particularly useful when the leading coefficient is 1 and the linear coefficient is even.

How do you complete the square for the equation x² + 6x = -7?

To complete the square for the equation $x^2+\; 6x=\; -7$, follow these steps:

1. Rewrite the equation in the form $x^2+\mathrm{bx}=c$.
2. Add ${\frac{b}{2}}^2$ to both sides. Here, $b$ is 6, so $\frac{6}{2}=\; 3$ and $3^2=\; 9$. Add 9 to both sides: $x^2+\; 6x+\; 9\; =\; -7\; +\; 9$.
3. Factor the left side: ${x+3}^2=\; 2$.
4. Use the square root property: $x+3=\; \pm \surd 2$.
5. Solve for $x$: $x=\; -3\; \pm \; \surd 2$.

When is completing the square a good method to use?

Completing the square is a good method to use when solving quadratic equations, especially when the leading coefficient (the coefficient of $x^2$) is 1 and the linear coefficient (the coefficient of $x$) is even. This method is always applicable, but it is particularly efficient in these cases because it simplifies the process of creating a perfect square trinomial, which can then be solved using the square root property.

What are the steps to complete the square?

The steps to complete the square are:

1. Rewrite the quadratic equation in the form $x^2+\mathrm{bx}=c$.
2. Add ${\frac{b}{2}}^2$ to both sides of the equation.
3. Factor the left side to form a perfect square trinomial: ${x+\frac{b}{2}}^2$.
4. Use the square root property to solve for $x$: $x+\frac{b}{2}=\; \pm \surd c$.
5. Solve for $x$ by isolating it on one side of the equation.

Why does completing the square work?

Completing the square works because it transforms a quadratic equation into a form that can be easily solved using the square root property. By adding ${\frac{b}{2}}^2$ to both sides, we create a perfect square trinomial, which factors into ${x+\frac{b}{2}}^2$. This allows us to apply the square root property, simplifying the process of solving for $x$. Essentially, it leverages the properties of perfect squares to make the equation more manageable.