Video: Quadratic Formula from Completing Squares

Write the equation π‘Žπ‘₯Β² + 𝑏π‘₯ + 𝑐 = 0, where π‘Ž β‰  0, in the form (π‘₯ βˆ’ 𝑝)Β² = π‘ž.

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Video Transcript

Write the equation π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero, where π‘Ž is not equal to zero, in the form π‘₯ minus 𝑝 squared equals π‘ž.

Here’s where we’re starting, π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐. And we’re trying to convert it into the form π‘₯ minus 𝑝 squared equals π‘ž. We’re going to need to do this by completing the square. If you look at this π‘₯ term, it doesn’t have a coefficient. And that means the first step we’re going to do is divide the left equation by π‘Ž. If we divide every term by π‘Ž, our new equation looks like this. π‘₯ squared plus 𝑏 over π‘Žπ‘₯ plus 𝑐 over π‘Ž 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 𝑐 over π‘Ž from both sides. π‘₯ squared plus 𝑏 over π‘Žπ‘₯ equals negative 𝑐 over π‘Ž. When we complete this square, we take this 𝑏 term, the coefficient of our π‘₯ to the first power, and we divide it by two. 𝑏 over π‘Ž divided by two, 𝑏 over π‘Ž divided by two is the same thing as 𝑏 over π‘Ž multiplied by one-half. And that is equal to 𝑏 over two π‘Ž.

We’ve now added 𝑏 over two π‘Ž squared to the left side of the equation. And that means we need to add 𝑏 over two π‘Ž squared to the right side of the equation. Because we’ve completed the square, we have π‘₯ plus 𝑏 over two π‘Ž squared on the left side. On the right side of the equation, we want to square 𝑏 over two π‘Ž. Negative 𝑐 over π‘Ž doesn’t change. We end up with 𝑏 squared over two squared times π‘Ž squared. Which we can rewrite as 𝑏 squared over four π‘Ž squared.

We’re almost done, but we know that we need one term on the left side. It needs to be π‘ž. And that means we need to take negative 𝑐 over π‘Ž plus 𝑏 squared over four π‘Ž squared and rewrite it as one term. We know that to add fractions we need a common denominator. And if we multiply negative 𝑐 over π‘Ž by four π‘Ž over four π‘Ž, we get a numerator of negative four π‘Žπ‘ and a denominator of four π‘Ž squared.

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

Now we need to note something here. By completing the square, we’ve ended up with addition here. Our question wants this in the format of π‘₯ minus 𝑝. If we wanted to do that, we could say π‘₯ minus negative 𝑏 over two π‘Ž. In that case, 𝑝 would be equal to negative 𝑏 over two π‘Ž. And π‘ž would be equal to 𝑏 squared minus four π‘Žπ‘ over four π‘Ž squared.

But our question hasn’t asked us to identify 𝑝 and π‘ž. It just wants this format. And so we’re gonna leave it as the simplified form π‘₯ plus 𝑏 over two π‘Ž squared equals 𝑏 squared minus four π‘Žπ‘ over four π‘Ž squared.

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