### Video Transcript

Write the equation ππ₯ squared plus ππ₯ plus π equals zero, where π is not equal to zero, in the form π₯ minus π squared equals π.

Hereβs where weβre starting, ππ₯ squared plus ππ₯ plus π. And weβre trying to convert it into the form π₯ minus π squared equals π. Weβre going to need to do this by completing the square. If you look at this π₯ term, it doesnβt have a coefficient. And that means the first step weβre going to do is divide the left equation by π. If we divide every term by π, our new equation looks like this. π₯ squared plus π over ππ₯ plus π over π equals zero.

To complete the square, we need to move this constant to the other side of the equation. And we do that by subtracting π over π from both sides. π₯ squared plus π over ππ₯ equals negative π over π. When we complete this square, we take this π term, the coefficient of our π₯ to the first power, and we divide it by two. π over π divided by two, π over π divided by two is the same thing as π over π multiplied by one-half. And that is equal to π over two π.

Weβve now added π over two π squared to the left side of the equation. And that means we need to add π over two π squared to the right side of the equation. Because weβve completed the square, we have π₯ plus π over two π squared on the left side. On the right side of the equation, we want to square π over two π. Negative π over π doesnβt change. We end up with π squared over two squared times π squared. Which we can rewrite as π squared over four π squared.

Weβre almost done, but we know that we need one term on the left side. It needs to be π. And that means we need to take negative π over π plus π squared over four π squared and rewrite it as one term. We know that to add fractions we need a common denominator. And if we multiply negative π over π by four π over four π, we get a numerator of negative four ππ and a denominator of four π squared.

Once these two fractions have a common denominator, we can add their numerator. Negative four ππ plus π squared can be rewritten to say π squared minus four ππ all over four π squared. We wanna substitute π squared minus four ππ over four π squared in, like this.

Now we need to note something here. By completing the square, weβve ended up with addition here. Our question wants this in the format of π₯ minus π. If we wanted to do that, we could say π₯ minus negative π over two π. In that case, π would be equal to negative π over two π. And π would be equal to π squared minus four ππ over four π squared.

But our question hasnβt asked us to identify π and π. It just wants this format. And so weβre gonna leave it as the simplified form π₯ plus π over two π squared equals π squared minus four ππ over four π squared.