Work each problem. Show that -2+i is a solution of the equatio...

Question

Work each problem. Show that -2+i is a solution of the equation x²+4x+5=0.

Accepted Answer

Hello everyone in this video we're going to be looking at this practice problem where we want to find the square root of negative I. So the square root of negative I looks like a radical with a negative I. Inside. And from this you can see that we're working with a complex number because of this imaginary component and also the square root. And because I is equal to the square root of negative one, this can be a difficult square root to do. However, we can make it easier by comparing it to a the standard form for a complex number because we know that we're going to be dealing with a complex number because of that imaginary component. So recall that the standard form for a complex number is A plus B. I. Where normally the B component is associated with the imaginary component and the component is associated with the real component. However, we're going to solve for negative I and see what we get. So I'll start by squaring both sides. So that gives me negative I is equal to A plus B. I times A plus B I. And I'll go ahead and do the foil method on the right hand side. So eight times A is a squared A times B I. Is a B I B I times A. Is again A B. I and B. I times B. I. Is B squared I squared. Well I square it is equal to the square root of negative one, times the square root of negative one, which gives back negative one. So I can replace that I squared with negative one. So that becomes A squared plus I combine these middle terms to a B. I minus b squared. Now you can see that this be no longer has the imaginary component. So it can be part of the real component for this complex number. So if I group the real components together and the imaginary components together I have a squared minus B squared and the real components followed with The imaginary components of two a. b. i. And on the left hand side I still have the negative I. And from this you might think that there's no real component. But recall that we can always add a zero. So now the zero is the real component of the left hand side and the imaginary component is the negative I. So now we can compare the real components on the left hand side with the real components on the right hand side, along with the imaginary components on the left hand side with the imaginary components on the right hand side. So if I do that, I have a squared minus B squared is equal to zero and to a B. I is equal to negative I so recall that I'm trying to figure out what A and B are equal to in this square root of negative eye. So I'm going to try to solve for a in the real section here. So I'll add B squared. So that gives me a squared is equal to B squared. And if I take the square root I have a is equal to plus or -7. So now I know that they could either be the same as B, or they're going to be opposites. And I'll go ahead and try and simplify this imaginary component. So I'll start by dividing by eye because both sides have I. So that leaves me with two. A B is equal to negative I over, I is equal to negative one. And from this you can see that too is positive and we get a negative answer back. So that's telling me that A and B must be opposites because if I had A and B being the same and they were both negative, I'd get back a positive. So the only way that I can be getting negative one is if A and B are opposites. So that is telling me that E. Is actually equal to negative B. So knowing this, I can substitute negative be in for A in this equation. So that gives me two times negative B times B is equal to negative one. So negative two B squared is equal to negative one. There's a negative on both sides. So I can eliminate it by dividing by negative one. So that gives me two B squared is equal to one and dividing by two to solve for B, B squared is equal to one half. If I take the square root, I finally get that B is equal to plus or minus the square root of 1/2. And knowing B I can solve for A. So when B is equal to Positive square root of 1/ A is equal to the opposite of that, which would be negative square root of one half. And when B is equal to negative square root of one half A is again the opposite, so positive square root of one half. And from these we can do a little more simplification because the square root of one is one. So we can leave the radical in the denominator. So it would be -1 divided by radical two. And we can do that for all of these Because the square root of one is 1 and there's no Perfect Square root for two. So we can just leave it as radical two in the denominator but now we have our coordinates and we can plug them back into the form of A plus B. I. So if I do that one of my coordinates starting with this one first Will be -1 divided by radical two Plus one Divided by Radical two. I or it could be the second option which is positive one divided by radical two minus one, divided by radical two. I. So this becomes our final answer for the roots of negative I and you can see that although we didn't really no the actual number for it. We figured it out by comparing it to the complex number standard form of A plus B I, and answer choice B also has the same conclusion that we arrived at. So hopefully this video helped and see you next time.

Work each problem. Show that -2+i is a solution of the equation x²+4x+5=0.

Key Concepts

Complex Numbers

Substitution Method

Quadratic Equations and Roots

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