Solve each system. y^2=5 x^2+1 x y=2

Name: Problem 26, Chapter 11: Precalculus: Graphs and Models
Uploaded: 2020-06-23T22:12:11-08:00
Duration: 8 min 27 s
Channel: Michael Jacobsen
Description: Solve each system. y^2=5 x^2+1 x y=2

Key Concepts

Quadratic Equations

Quadratic equations are second-degree polynomial equations commonly represented in the form ax² + bx + c = 0. Familiar methods such as factoring, completing the square, or using the quadratic formula are used to solve them. In the context of solving systems, transforming an equation into a quadratic form can significantly simplify the process of finding solutions.

Nonlinear Systems of Equations

This concept involves dealing with systems where one or more equations are nonlinear, meaning they include variables raised to powers other than one or involve functions like squares or products. Solving these systems requires methods that can manage the complexities introduced by the nonlinear terms, and they may yield multiple, single, or no solutions depending on the nature of the system.

Substitution Method

The substitution method is a technique used to reduce the number of variables in a system of equations by expressing one variable in terms of the other using one of the equations, and then substituting into the remaining equation(s). This is especially useful in nonlinear systems where directly eliminating variables may be challenging.

Transcript

00:01 For this system of equations, one feature that's noteworthy is that we have two variables being multiplied together, resulting in a non -zero constant in our second equation.

00:12 This is a special equation to have since it immediately implies that x cannot be zero and y cannot be zero.

00:19 This allows us to solve for either variable x and y.

00:23 If we divide by, say, x on both sides of this equation, we obtain that y is 2 divide by x.

00:30 Now we know that this can never be undefined when we allow x to be a solution, and so we're not going to miss out on any solutions to this equation.

00:40 Next, we're going to substitute y is 2 over x into this portion of the equation we see here.

00:46 So we'll have y to the power of 2, which i'll leave as a blank for the moment, equals 5x squared plus 1.

00:56 But we just found that y itself is equal to 2 over x let's simplify the left -hand side to obtain 4 over x squared is equal to 5x squared plus 1 now when we solve this kind of equation the x squared in the denominator is of our major concern at this step so we're going to take an extra step to multiply both sides by x squared in order to clear that denominator on the left on that side we obtain just a 4, and on the right hand side we'll distribute in x squared to obtain 5x to the power 4 plus x squared.

01:38 To solve this equation, let's subtract 4 from both sides so that we have 0 on one side, which is now equal to 5x to the power 4 plus x squared minus 4.

01:54 To solve this equation, we could use the quadratic formula, but first let's try to obtain a solution through factor since that's often a faster route.

02:04 We'll write 0 is equal to one group times another group, and the first step is to factor that 5x to the power 4.

02:12 Well since 5 is prime, we can go with a 5 here, x squared, and x squared.

02:19 Then if we foil our answer so far, we will obtain 5x to the power 4.

02:23 So the first term here has been determined.

02:27 Next, let's analyze the signs.

02:30 A negative sign exactly here means our result will be a positive or negative.

02:35 Let's go with a positive here and a negative there.

02:40 Then the next step in this solution will be to factor 4 and put it in the appropriate places.

02:46 Let's try a factorization that involves a 4 here with a 1 there.

02:52 In order to check this equation, let's look at two multiplications.

02:57 We know that if we multiply here to here we get 5x to the power of 2.

03:02 And when we multiply here to here, we get negative 4 x to the power of 2.

03:07 Both of these quantities that we just found sum together to get positive x squared, so that tells us we have a check for this factorization, and this term here is also determined.

03:21 Since the factorization is complete, and since we have zero on the left -hand side, our next step is that we can conclude 5x squared minus 4, equal 0 or x squared plus 1 is equal to 0.

03:38 Now we can solve both of these equations for x.

03:42 Starting with the first equation, 5x squared is now equal to positive 4, divide by 5, and x squared will be equal to 4 over 5.

03:53 If we take the square root of both sides of this equation, then x will be equal to positive or negative 2 over the square root.

04:02 To 5...

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