Having this new knowledge allows us to explore one more possible outcome when we solve quadratic equations. Consider this equation:. Using the Quadratic Formula to solve this equation, we first identify a, b, and c. We can place a, b and c into the quadratic formula and simplify to get the following result:. But, now that we have defined the square root of a negative number, we can also define a solution to this equation as follows. In this section we will practice simplifying and writing solutions to quadratic equations that are complex.
We will then present a technique for classifying whether the solution s to a quadratic equation will be complex, and how many solutions there will be. In our first example we will work through the process of solving a quadratic equation with complex solutions.
Take note that we be simplifying complex numbers — so if you need a review of how to rewrite the square root of a negative number as an imaginary number, now is a good time. Simplify the radical, but notice that the number under the radical symbol is negative! There are two x -intercepts. If the discriminant is a perfect square, the two roots are rational numbers. If the discriminant is not a perfect square, the two roots are irrational numbers containing a radical not an " i ".
There is one x -intercept. The root is repeated. There are no x -intercepts. When graphing, if the vertex of the quadratic function lies above the x -axis, and the parabola opens upward, there will be NO x- intercepts and no real roots to the equation. The equation will have complex conjugate roots. The same applies if the vertex lies below the x -axis, and opens down. For calculator help with quadratic graphs click here. NOTE: The re-posting of materials in part or whole from this site to the Internet is copyright violation and is not considered "fair use" for educators.
This 2 and this 2 are going to cancel out. So on the left hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works.
Now let's try 3 minus i. So once again, just looking at the original equation, 2x squared plus 5 is equal to 6x. Let me write it down over here. Let me rewrite the original equation. We have 2x squared plus 5 is equal to 6x. And now we're going to try this root, verify that it works. So we have 2 times 3 minus i over 2 squared plus 5 needs to be equal to 6 times this business.
Once again, a little hairy. But as long as we do everything, we put our head down and focus on it, we should be able to get the right result. So 3 minus i squared. And then you're going to have two of those.
So negative 6i. So negative i squared is also negative 1. That's negative 1 times negative 1 times i times i. So that's also negative 1. Negative i squared is also equal to negative 1. Negative i is also another square root. Not the principal square root, but one of the square roots of negative 1. So now we're going to have a plus 1, because-- oh, sorry, we're going to have a minus 1. Because this is negative i squared, which is negative 1. And all of that over 4. All of that over-- that's 2 squared is 4.
Times 2 over here, plus 5, needs to be equal to-- well, before I even multiply it out, we could divide the numerator and the denominator by 2. So 6 divided by 2 is 3. So 3 times 3 is 9. And if we simplify it a little bit more, 9 minus 1 is going to be-- I'll do this in blue. We have 8 minus 6i. And then if we divide 8 minus 6i by 2 and 4 by 2, in the numerator, we're going to get 4 minus 3i. And in the denominator over here, we're going to get a 2.
We divided the numerator and the denominator by 2. Then we have a 2 out here. And we have a 2 in the denominator. Those two characters will cancel out. And so this expression right over here cancels or simplifies to 4 minus 3i. Then we have a plus 5 needs to be equal to 9 minus 3i.
I We have a negative 3i on the left, a negative 3i on the right. We have a 4 plus 5. We could evaluate it. This left hand side is 9 minus 3i, which is the exact same complex number as we have on the right hand side, 9 minus 3i. So it also checks out.
It is also a root.
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