Findteh Minimumor Maximum Values - Example 3
Graph Quadratic Functions - Example 3
Graph Quadratic Inequalities - Example 3
Solve Quadratic Equations Algebraically - Example 3
Solve Quadratic Equationsby Factoring - Example 3
Liberty University
Dividing Complex Numbers - Example 3


Comments

Comments are currently disabled.

Video Transcript

flex numbers to minus I divided by five plus two I. And when I do, we're gonna do that by multiplying the denominator conjugated the denominator which would be five minus two I. So we'll multiply both numerator and denominator by five minus two I So in orderto multiply are enumerators. We're gonna have to use full. So we're gonna have two times five is 10, two times negative to I is negative. Four. I negative Atom's five. I is going to be negative. Five I, in our last terms was negative. I times negative to I would be, too I square. So we're gonna go ahead and write these out. We'll simplify. We will combine like terms. So let's be 10 minus not I plus two. I squared over Let's do full again. But this time for our bottom number So we'll have five times five is 25 five times negative. Two eyes negative. 10 I five times to I is 10. I these two, we're gonna cancel out already because the same number with the opposite sons and two times negative two is going to be negative. Four. I squared so we'd have 25 minus four I square. So let's go ahead and simplify all of these. So let's go and change your eye squares to negative one, because we know that's what they equal. So 10 minus non I plus two times negative one over 25 minus four times negative one that's going to simplify to 10 minus not I minus two over 25 plus four. When we combine like terms, that gives me 10 minus nine. I'm sorry, that's actually going to give me eight minus 99 over 29. So now we can split this into two separate fractions, so we'd have eight over 29 minus 9/29. I normally you would simplify your fractions, but both of these fractions are in lowest terms, so there's my solution.