Video: Solving Quadratic Equations by Completing Squares

By completing the square, solve the equation π‘₯Β² βˆ’ 2√(3) + 1 = 0.

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Video Transcript

By completing the square, solve the equation π‘₯ squared minus two times the square root of three plus one equals zero.

Before we work on this problem, I wanna talk about completing the square in general. If we have an equation in the standard form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero, here are the steps we follow to complete the square. First, we divide everything by π‘Ž. We want the coefficient in front of π‘₯ squared to be one. Then we would have the equation π‘₯ squared plus 𝑏 over π‘Ž π‘₯ plus 𝑐 over π‘Ž.

After that, we need to move this constant, 𝑐 over π‘Ž, to the other side of the equation. We then have π‘₯ squared plus 𝑏 over π‘Ž π‘₯ equals negative 𝑐 over π‘Ž. Here’s where we start adding things to the equation. We need to add 𝑏 over π‘Ž divided by two squared to both sides of the equation. But we can say divided by two is the same thing as multiplying by one-half, which means we’re adding 𝑏 over two π‘Ž squared to both sides of the equation. It will now look like this. π‘₯ squared plus 𝑏 over π‘Ž π‘₯ plus 𝑏 over two π‘Ž squared equals 𝑏 over two π‘Ž squared minus 𝑐 π‘Ž.

The left side of the equation is now a square value. It is now π‘₯ plus 𝑏 over two π‘Ž squared. And the right side is 𝑏 over two π‘Ž squared minus 𝑐 π‘Ž. You can also simplify this side of the equation. If we square 𝑏 and two π‘Ž, we get 𝑏 squared over four π‘Ž squared minus 𝑐 π‘Ž. And if we multiply the 𝑐 over π‘Ž by four π‘Ž, in the numerator and the denominator, we can say 𝑏 squared minus four π‘Žπ‘ all over four π‘Ž squared, which we’ll substitute back into this part of the equation.

Why did we do all this? We did all of this because this equation will work no matter what values you have for π‘Ž, 𝑏, and 𝑐. Just to note, there should be an π‘₯ by the negative two times the square root of three there. The equation we’re solving for is π‘₯ squared minus two times the square root of three π‘₯ plus one equals zero. And our π‘Ž value is one. The coefficient of π‘₯ is just one. If π‘Ž equals one, the 𝑏 value, the coefficient of the π‘₯ to the first power, is negative two times the square root of three. 𝑏 equals negative two times the square root of three.

We also need to know what 𝑏 over two is. 𝑏 over two equals negative two times the square root of three over two, which we can simplify. The twos cancel out. 𝑏 over two is equal to the negative square root of three. And 𝑏 over two squared equals the negative square root of three squared, which is three. And our 𝑐 value equals one. Remember, we move the 𝑐 value to the other side of the equation. Then we add 𝑏 over two squared to both sides. And our 𝑏 over two squared equals positive three. Negative one plus three equals two. The left side is a square. And we rewrite that as π‘₯ plus 𝑏 over two squared. Our 𝑏 over two is the negative square root of three. We can simply say π‘₯ minus the square root of three squared.

When our goal is to solve the equation, we want to find out what π‘₯ values make this statement true. And we’ll do that by isolating π‘₯. That means we’ll need to take the square root of both sides of the equation. And that tells us that π‘₯ minus the square root of three equals the square root of two.

Here we’re gonna break the equation up into two statements, because we remember that the square root of two has both a positive and a negative solution. We’ll have π‘₯ minus the square root of three equals the positive square root of two. And π‘₯ minus the square root of three equals the negative square root of two. To get π‘₯ by itself, we’ll add the square root of three to both sides of both equations. π‘₯ equals the square root of three plus the square root of two. Or π‘₯ equals the square root of three minus the square root of two.

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