Hi. My name is Anthony Varela. And today, I'm going to be solving quadratic equations with no real solutions. So we're going to be using the quadratic formula.
And specifically, we're going to look at the sign of the discriminate, whether it's positive or negative. That's going to help us determine if there are real solutions or if there are no real solutions. And then we're going to talk about those non-real solutions and talk about how we can express to those.
So recall that the quadratic formula is x equals negative b plus or minus the square root of b squared minus 4ac all over 2a. And we get the values for a, b, and c from our quadratic equation in standard form, ax squared plus bx plus c set equal to 0. That's how we substitute values and for a, b, and c.
And as we go through simplifying our quadratic formula we're going to focus on the discriminate. This is the value underneath the square root, b squared minus 4ac. And in particular, if that value is a non-negative number-- so it's greater than or equal to 0, that means that there is at least one real solution to the quadratic. That means there's at least one value for x that makes y equals 0.
Now if the discriminate is negative or less than 0, we have no real solutions. And that's because we cannot evaluate the square root of a negative number using our set of real numbers. There is no number squared that can be negative, but there is a way to express this using i, which is the square root of negative 1.
And this involves than imaginary numbers. So an imaginary number is a non-real number, which is a multiple of i, the square root of negative 1. So we could have 2i, 3i, 4i, so on and so forth. Those would make up imaginary numbers.
So let's practice then writing the square root of a negative number using i. So here we have the square root of negative 7. Well, I can write this as 7 times negative all underneath one radical sign.
I can split this up into two square roots as the square root of 7 times the square root of negative 1. Now I can write the square root of negative 1 using i. So this would be the square root of 7 times i. We no longer have any visible negative numbers, because i equals the square root of negative 1.
Let's try another one. We have the square root of negative 9. I can write this as 9 times negative 1 underneath one radical.
I can split this up into two square root signs, square root of 9 times the square root of negative 1. And I can write square root of negative 1 using i. And I can also write the square of 9 as 3. So this would be 3i.
Big idea here is that looking at the sign of the discriminate, so that expression b squared minus 4ac, will tell you if you have a real solution or not. If b squared minus 4ac is non-negative, so at 0 or greater, we have at least one real solution. If it's less than 0, if it's negative, we have no real solutions.
Our solutions are going to be non-real involving that i value and the square root of negative 1. So now we're going to go through an example of solving a quadratic equation that has no real solution. So here's the equation 0 equals x squared minus 2x plus 2.
Now here's a graph of y equals x squared minus 2x plus 2. And one of the things that we can look for graphically to determine if we have a real solution or not is look for x-intercepts. Now we see that this particular graph does not have any x-intercepts. So that means there are no x values that make y equals 0.
So that's a clue that we have no real solutions based on the graph, but let's go ahead and show this using our quadratic formula. So to find values for a, b, and c, we can see that a equals 1, b, equals negative 2, and c equals positive 2.
So plugging those values into our quadratic formula, we have x equals 2 plus or minus the square root of 4 minus 4 times 1 times 2 all over 2a, which in this case is 2. So now looking at what's underneath our square root sign, we have 4 minus 4 times 1 times 2. Well, 4 times 1 times 2 is 8.
Now 4 minus 8 is negative 4. So here we see a negative number underneath that square root. So we know that we have no real solutions to this equation, but how can we express our non-real solutions then? Well, I can rewrite the square root of negative 4 using i, which is the square root of negative 1.
So I'm going to write this then as 4 times negative 1. And now I can split this up into two radicals, the square root of 4 times i. The square root of 4 is 2. So this would be x equals 2 plus or minus 2i.
Now notice that I can divide both 2 and 2i by 2. So I have x equals 1 plus or minus i. So one solution would be x equals 1 plus i. And another solution would be x equals 1 minus i.
And this is how we would express our non-real solutions. They're actually complex numbers. So a complex number is a number a plus bi containing a real component and an imaginary component where i is the imaginary unit square root of negative 1.
So let's review our lesson on quadratic equations with no real solutions. Well, we used our quadratic formula x equals negative b plus or minus the square root of b squared minus 4ac all over 2a. And in particular, we focused on that discriminate b squared minus 4ac. If it's not negative, we have at least one real solution. But if it is negative, that's when we have no real solutions.
Graphically, quadratic equations with no real solutions have no x-intercepts. Algebraically, it's when we have that negative number underneath the square root sign. And our solutions are complex numbers having a real part and an imaginary part with this imaginary unit i, which is a square root of negative 1. So thanks for watching this tutorial on quadratic equations with no real solutions. Hope to see you next time.