Solve each nonlinear system of equations. Give all solutions, inc...

Question

Solve each nonlinear system of equations. Give all solutions, including those with nonreal complex components. See Examples 1–5.3x^2 - y^2 = 11 xy = 12

Accepted Answer

Hello everybody. I hope you're doing right today. Today we're going to be looking at this math question that states provide all solutions for the given nonlinear system of equations including non-real complex components. Where the system of equations includes five X squared minus Y squared equals 19 and X Y equals two. Now, the institutes provided are a negative I divided by the square to five comma two, I multiplying the square to five, I divided by the square to five comma negative two Y multiplying the square to 51 comma two and negative one comma negative two B has the first two terms, the same as A and the last two terms being negative two comma one and two comma negative one C is two, I multiplying the square to five comma negative I divided by the square to five negative two. I multiplying the square to five comma I divided by the square to 52 comma one and negative two comma negative one and D has the first, the same first two terms as A and the same last two terms as C. Now, when looking at a question like this, I see that we have two equations with the same variables where I can use one equation to solve for one variable and plug that into the other equation. So in this case, I'm gonna be using X Y equals two to solve for Y because it's the easier equation out of the two. So if X Y equals two, then Y will be equal to two divided by X. And now I can use that information to plug that into the other equation. So we'll have five X squared minus and we had Y squared, which in this case was two divided by X. And we'll have that squared and that's equal to 19. Now, when we have a fraction squared, we know we have to square both the numerator and the denominator. So we'll do exactly that and we'll have five X squared minus the numerator squared, which is two squared, which is four divided by X squared, which is X squared. And that is equal to 19. Now to get rid of the X squared and the denominator, we'll multiply everything by X squared. So we'll have five X to the fourth minus four equals 19 X squared. And now we can move that 19 X squared over from the right hand side to the left hand side and we'll have five X to the fourth minus 19 X squared minus four equals zero. And now we have this in more of a way where we know how to factor. So we have, we have the setup to factor. However, it's not what we usually see. So we have a coefficient of that's, that's greater than one and we have exponents greater than X squared. So you could technically factor this from right here. However, I will be taking a different route and using use substitution which will decrease our exponent and will make it a little bit easier to work with. So in this case, we're going to let X squared equal you and that will turn our equation into five U squared minus U minus four equals zero. So as you can see, our exponents became a little smaller and this is kind of more what we see when we're factoring usually. So now I'm going to walk a little bit slowly how we're going to factor this. So we see that our coefficient or the number in front of the U squared term is not one, it's five. So we have to multiply that five with our number terms. So the last term not attached to a U, which is in this case is a negative four. So five multiplied by negative four is negative 20. And now we need to see is there are there two numbers that when multiplied by each other will give me negative 20 but when added to each other will give me negative 19. Well, yes, there are, there are two numbers like like that it'll be negative 20 and positive one. So negative 20 multiplied by one is equal to negative and negative 20 plus positive one is equal to negative 19. And why is it negative 19? Well, we need, we need to add up to the term in the middle of our equation. So we had five U squared minus 19 U. So we need to add up to that to the number that's in front of A U, the U that's not squared, no exponent, nothing like that. So in this case, that number was ne negative 19. So that's what we add up to negative 19. So now I can rewrite our equation as five U squared. And instead of writing negative 19, I can write minus 20 U plus U which is the same as minus 19 U and then minus four equals zero. And now I can do some grouping to factor. So I'll group the first two terms. So I'll have in one parentheses, five U squared minus 20 U and close that parentheses. And then we'll have plus in our second group or the last two terms U minus four in the second parentheses. And now we can factor this a bit, especially the first parentheses. So we have five U squared minus 20 U, I can factor out a five U which will leave a U minus four inside of that parentheses. And then we'll have plus and our goal here is to match both parentheses. So we have U minus four and one parentheses. And we wanna get both U minus parentheses to have U minus four. So to do that, I'll just factor out a positive one which will leave U minus four inside of the second parentheses. And now we can continue grouping so we can group in one parentheses, the numbers multiplying the parentheses. So we have five U plus one group and that parentheses that will now be multiplying the U minus four, which is what we wanted and that'll be equal to zero. So now we have five U plus one in one parentheses, multiplying U minus four in another parentheses. And that'll be equal to zero. And now we can solve for you. So we'll have five U plus one, equal zero and U minus four equals zero. And we do this because as long as either of the parentheses is equal to zero, since they're multiplying each other, we know that the final answer will be zero. So we'll have five, U is equal to negative one or U is equal to negative 1/5. And for the other U, we had U minus four equals zero. So U equals positive four. And now I'm just gonna move my work a bit to have more space. So we said earlier that X squared equals U and now we have some U terms. So we have that X squared is equal to negative 1/5 or four and to get rid of the square. So just to have X, we take the square root of that, so we'll have that X is equal to the square root of negative 1/5 or the square of four. So let's solve this. When we have the square root of negative 1/5 we see that there's a negative inside of the square root, which means we're going to be working with imaginary numbers or the letter I. So we'll have X is equal to. So we're taking, well, when we have a fraction inside of a square root, we can take the square root of the numerator and the square root of the denominator, the square root of negative one. In this case will put I which is imaginary and then the square root of five, which is the, the denominator will remain as the squared of five since we can't simplify that. And we know that when you take the square root of something, it'll, the answer will be plus or minus. So we'll have I divided by the square to five and will have negative I divided by the square to five. And then we had the or which was the square of four. So X is equal to. When we have the score of four, we have, we can have answers of positive two or negative two because the squared of four can be plus or minus two. So now we can plug this into our Y equation. We said earlier that Y Y of X will say is equal to two divided by X. So Y of positive I divided by the square root of five is equal to two divided by I divided by the square of five, which is equal to negative two, I multiplying the square of five. So all we did now is just plug in one of our XS to our Y equation. And now we can continue to do that for all our axes. So or plug in Y of negative I divided by the square of five, that will be equal to two divided by negative I divided by the square of five. And now that will be equal to positive to I multiplying the square of five. And again, I'll move my work a bit to have more space. So now we'll continue to plug in and we'll have Y of positive two is equal to two divided by two, which is one and Y of negative two, which is two divided by negative two, which is negative one. And now we have ordered pairs. So we have X values and the Y values for each respective X. So let's write those pairs or a, a ordered pairs. So we'll have when our X is I divided by the square root of five R Y was negative two, I multiplying square to five. When our X was negative I divided by the square of five R Y was two, I multiplying the square to five when her ex was two, R Y was one and when our X was negative two, our Y was negative one. So those are the ordered pairs for this question. And we just have to compare those to the answer choices. And we see that those order pairs will match up with answer choice D so negative I divided by the square of five comma two, I multiplying square to five, I divided by the square of five comma negative two, I multiplying the square of 52 comma one and negative two comma negative one. So I hope that was helpful. And until next time.

Solve each nonlinear system of equations. Give all solutions, including those with nonreal complex components. See Examples 1–5.3x^2 - y^2 = 11 xy = 12

Key Concepts

Nonlinear Equations

Systems of Equations

Complex Numbers

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