We want to take a look at the solution to some quadratic equations where we solve them by what's called completing the square.
Here's our first one: we have (5x + 2)² = -8. Now, to solve this equation right here,
we say that the quantity (5x + 2)² = -8, then 5x + 2 must be either the positive or the negative square root of -8.
So, we go from this expression down to this expression, and that gets rid of that square that we have on the 5x + 2.
So, (5x + 2)² = -8, then 5x + 2 itself is either the positive or negative square root of -8.
Now, we'll simplify this: 5x + 2 = ±, when I simplify this expression I get 2i√(2).
So, the negative side under the radical gives me an i outside, and 8 = 4(2), and √(4) = 2; so, I have 2i√(2).
Now, I'll simply add -2 to both sides, and then divide both sides by 5; that will give me x = -2, ± 2i√(2),
all divided by 5. And these are my two solutions to this equation right here:
[-2 + 2i√(2)] /5; and then the other solution is [-2 - 2i√(2)] /5.
Let's look at our next problem: a² + 10a + 22 = 0.
Now, what I want to do is get this equation right here in the form of the equation we solved in the problem number 1.
So, I'm going to start by writing these first two terms together on the left side, and moving the constant term to the other side,
by adding -22 to both sides. So, I get -22 on the right side.
Now, I look at these first two terms as if they're the beginning of a perfect square trinomial: I have 1a² + 10a, plus—
now I'm going to add on my own third term right here, so that this becomes a perfect square.
So, I take half this coefficient; half of 10 is 5, and I square it—that gives me 25. If I add 25 to one side, I also add 25 to the other side.
The reason I did that is so that I can write this expression as (a + 5)² = -22 + 25, is 3.
Now, I'm down to an equation that has the form of the one in problem number 1; so, (a + 5)² = 3
that means that (a + 5) = ± √(3).
Now, I add -5 to both sides, and I get a = -5 ± √(3).
So, this is what the two solutions look like to this equation right here, and I found them by a process called completing the square:
move the constant term to the other side, then I take half the coefficient of the linear term, square it, and add it to both sides.
That guarantees that this side is a perfect square trinomial, and then I simply apply this theorem here,
that says that if (a + 5)² = 3, then (a + 5) itself is either the positive or negative square √(3), that allows me to solve the equation.
Let's look at our next example. This time I have 2x² - 4x - 8 = 0. I need to do two things.
I need to get rid of this constant term on this side, and I also need to make this coefficient of x² 1.
So I'm going to divide both sides by 2, and then I'm also going to add the opposite of this to both sides.
When I do that, I'll end up with x² - 2x = 4. So, I divided everything by 2; that gives me x² - 2x - 4;
and then, I add positive 4 to both sides—so I have this. Now, I'm going to complete the square by taking half the coefficient of this term;
so, -2: half of that is 1; (1)² = 1. So, I'll add positive 1 to both sides of this equation.
On the left side, then, I have guaranteed the perfect square: (x - 1)²; and on the right side, I have 5.
Now if (x - 1)² = 5, then x - 1 itself is either the positive or negative square root of 5.
Add 1 to both sides, and I have 1 ± √(5).
So, there's my two solutions to this equation right here, and I found them by this process called completing the square.
It requires that I make the coefficient of the squared term 1, move the constant term to the other side, and then complete the square
by taking half the coefficient of the middle term, squaring it, and adding it to both sides.