ma006 ยป

Let's learn one last method to help us solve what may look like a quadratic equation. Let's look at the equation x to the fourth minus 13x squared plus 36 equals 0. This equation is considered to be a pseudo-quadratic equation, or a false one. It almost appears to be quadratic, but we really have a fourth power here. Notice too that we have no odd powers of x. We just have x to the fourth, x squared and then no x term, just a constant here. We can change this equation into a quadratic equation that we can potentially solve. This part would be quadratic, since we would have something squared minus another something plus a constant. Notice how the powers on any variable term would be 2 then 1 and then repeated expression here. When we have an expression that's repeated, we can make a substitution for that expression and replace it so we have a simpler equation to solve. So what do you think is that repeated expression? What should we place inside these parentheses?

This pseudo quadratic equation can be written as the quantity x squared squared minus 13 times x squared plus 36. It turns out that the repeated expression is x squared. We'll use that here, and we'll use that here. We know this since x squared to the 1st power is just x squared. And x squared to the 2nd power is x to the 4th. Nice thinking if you found these two expressions.

Now, we can use this equation and make a substitution. We can replace the quantity x squared with a different variable that we'll call u. So, here instead of having x squared, we'll have the value u and we'll square it. Here we'll have minus 13 times u to the first power. Again, here replacing x squared with our substitution of u. We're turning our non quadratic equation into one that is quadratic. So now that we've got this quadratic, let's see if we can solve it. What do you think are the solutions for this equation. Keep in mind that you're solving this equation for u. We aren't solving for x, which was the original variable in our original equation. We'll come back to that later. When you have the answers for u, enter them here. Good luck.

Here the answers are positive 9 and positive 4. Great work for factoring, if you found these two. We factor this equation by finding factors of positive 36 that sum to negative 13. Those two factors are negative 9 and negative 4. We set each factor equal to 0 to get u could equal 9 or u could equal positive 4.

## Solve for the Original Variable

As I mentioned before, we actually solved this equation for the values of u. But we're really interested in the values of x. We want the values that make our original equation true, not this middle one containing the variables of u. So since we've performed this step of a u substitution here, we want to turn these u values back into x values. I'll start by listing my answers, u equals 9 and u equals 4. Now, we can substitute this value of u with x squared. This allows us to get the x values that make the original equation true. We simply just sub each of these for x. Now that we've made the substitution for u, what do you think the values of x could be? When you think you have the answers, write it as a solution set here. Use commas in between each response. Good luck.

## Solve for the Original Variable

To solve each of these equations, we need to take the square root of both sides of the equations. When we do this, we need to include the positive and negative sign, since we can have either the positive root or the negative root. So x can equal plus or minus 3, or x could equal plus or minus 2. Notice that we really have four solutions to the original equation. This should also make sense since the degree here was a 4. We'll have at most four solutions to this equation, since the highest power is a 4.

## U substitution

This entire process is called U-Substitution. It's when we can take an equation an we make a substitution, to turn it in to our quadratic equation. Here we've replaced the repeated expression x squared, with the variable u. We created this so we could perform the substitution. Now we could have made any assignment for x squared. We didn't have to use the variable u. We could call this t substitution and use t or w substitution and use w. It doesn't matter what variable you use here you just want to make sure you use the same one throughout your problem solving. Keep in mind that we performed this substitution so we could have a factorable quadratic. We factor this expression to solve for the values of u. Keep in mind that these values are not our final answer. We need to find the original values of the original variable, which was x. So we take our answers, u equals 9, and u equals 4, and we replace u with the original variable that we substituted, x squared. We solved for the original variable which is x, which gave us our four solutions. Now if you're not convinced that each of these is a solution, try checking them in the original equation. You'll find that all four of them check.

## Finding the Substitution

Here's another pseudo-quadratic equation. Notice that we have something squared, then something raised to the 1 power. What expression should u equal in order for us to do u substitution here?

## Finding the Substitution

In this case, we would want u to equal 4m minus 3. Great thinking if you found it. In general, when we choose u for u substitution, we want u to equal the repeated expression in the equation. Notice that the 4m minus 3 is a repeated expression. We have it squared, and then we have it raised to the 1 power.

## Making the Substitution and Solving for u

Now that we have the corrects substitution to use, we can replace 4m minus 3 with the variable u. So, this part becomes u, and this part becomes u. So, we'll have 2u squared plus 7u plus 5 equals 0. Now that we have a quadratic equation to solve, I want you to find the factored expression for the left hand side of the equation. And then I want you to solve this equation for u. Be sure when you enter your answers here, you enter them as a set.

## Making the Substitution and Solving for u

The factored form for this expression is 2u plus 5 times u plus 1. And the solutions for u are negative 5 halves and negative 1. Now that's some great work if you got these two correct. To factor this quadratic, we find factors of term using these as coefficients. Then, we use factoring by grouping to get 2u plus 5 times u plus 1 equals 0. We set each factor equal to 0 and then we solve for the variable u. u can equal negative 5 halves or u can equal negative 1.

## Solve for u and the Original Variable

One of the key ideas is that we need to solve for the original variable. We solved for u here, but our equation started with m. We made the substitution u equals 4m minus 3, so we can use that to our advantage to find the value of n, based on our answers. So, I want you to use your answers of u equals negative 5 halves and u equals negative 1 to find the original values for m. What do you think they'd be? When you've got them, write your answer as a set.

## Solve for u and the Original Variable

The values for m are 1/8 and 1/2. Nice solving if you found these two fractions. We can replace u with what it equaled from our substitution 4m minus we add 3 to both sides of the equation to get 4m equals negative 5 halves plus together should be positive 1/2. If I haven't convinced you, then think about what we can change 3 into. We can change what this number looks like by writing it as a fraction. 3 is the same as 6 divided by 2. So, really we have negative we do negative halve. So, these added together will give us positive 2 halves. Finally, we divide by 4 on both sides of the equations to get m is equal to 1/2 times 1/4, and m is equal to 1/2. I know this is true, since, when we divide by a number, we can really multiply it by its reciporical. So, 1/2 divided by 4 is

## u Substitution Practice 1

Try doing your own U-substitution problem by solving this equation. What do you think the values of x equal? Be sure you don't give us the values of u. We want the original values to this equation. Good luck.

## u Substitution Practice 1

Here, the correct solutions are positive 3 and positive 6. That's amazing effort if you got these two solutions. Now, don't worry if you got stuck or don't quite understand this at all, this is only our first practice problem. Maybe you were able to get the substitution of X minus 2, or maybe even found the values of U. Let's go through the solution together, pause it where you think you made your first mistake, and then see if you can solve for these answers. To perform the U-substitution here, we're going to substitute this repeated expression, x minus 2, with the variable u. So we'll have u squared minus 5 times u plus 4. This is the critical step that turns this equation into a much easier quadratic equation to factor and solve. We factored this using u minus 4 times u minus 1, and then we solve for u. U can equal 4, or u can equal We use the substitution u equals x minus 2 so we need to plug in x minus 2 back in for u to find the original values of x. So if u equals 4 and u equals 1 then we know x minus 2 must equal 4... And we know x minus 2 must also equal 1. Now we just add 2 to both sides for each equation to get x equals 6 and x equals 3, our final solution set.

## u Substitution Practice 2

What about this equation? What do you think the solutions would be for x? Perform your u substitution first, find your values for u and then go back and find the values of x. I know you'll be able to do this one. Good luck.

## u Substitution Practice 2

Here the correct answers are negative five thirds and positive one third. Great work if you found these two. I know fractions aren't always the easiest to work with but let's see how we got these two answers. FIrst we'll let u equal x plus one. This will allow us to perform the u substitution in this step so we'll have 9 u squared mini s6 u minus 8 equals zero... Now that we have a quadratic expression, we want to find factors of negative 72, the A times the C, that sum to negative 6 are B value. Those factors are negative 12 and positive 6. So, I use these two to rewrite the middle term here. We use factoring by grouping to get 3u plus 2 times 3u minus 4 equals to 0. We set each of these factors equal to 0 so we'll have u can equal negative two thirds or u can equal positive four thirds. Now if you got this far and solved for these values of u great work, keep in mind though we need to solve for the values of x, the original variable from our equation. We use our two solutions for u, and the fact that u equals x plus 1, to solve for the original variable x. So, we'll have x plus 1 equals negative two thirds, and we'll have x plus 1 equals to four thirds. We subtract equal to negative two thirds minus 1. So,since I have negative two thirds and negative 1, I have negative 1 and thirds. This is the mixed number. If I change it to an improper fraction, I'd have negative five thirds. For the other value of x, we'll have x equals four thirds minus 1. When I subtract 1, I get to choose what this 1 looks like. Since I have thirds over here, I want to make sure that I subtract three thirds, a form of 1, over here. Four thirds minus three thirds equals one third, and here are our two solutions: negative five thirds and one third.

## u Substitution Practice 3

How about this equation? What do you think the solutions would be for m? Keep in mind that we have a 4th power here. What do you think this means for our solution set? As a hint, this is like one of the problems we've done before. Good luck here.

## u Substitution Practice 3

The solution set includes negative 3, negative 1, positive 3 and positive 1. Four separate answers. Great work if you found these four. And keep in mind, we should have these four answers, since we have a 4 for our highest power. We can make the substitution u to equal m squared. We know this since m squared squared equals m to the fourth. This lets us have a power of two here and a power of 1 here on the us. Next, we factor this quadratic expression to get u minus 9 times u minus 1. We set each of these factors equal to zero to get u equals 9 and u equals positive 1. Now we're ready to go back and find the original values for m. We use u equals 9 and u equals 1 from the two solutions we found from factoring. We know u really equals m squared. So for this u, we'll have m squared equals 9 and for this u, we'll have m squared equals 1. We want to solve for m, so we take the square root of both sides of the equations. So here we'll have m equals plus or minus 3 and here we'll have m equal to plus or minus 1. So we'll have negative 3, positive 3, negative 1 and positive 1 as our solution set.

## u Substitution Practice 4

Here's our fourth problem on u substitution. What do you think the values are for b?

## u Substitution Practice 4

For this equation, the final solution was negative 5 and negative 14 5ths. Both of these values work for b. Great work if you found these two answers. I know one of them was a fraction, and you might not have gotten all the way there. But I hope you were able to make the substitution for b plus 3 and at least find the values for u. Let's see how we do that. We can start by making the substitution u equals b plus 3. So wherever I see b plus 3, replace that with a u. Now we want to solve this quadratic equation, so we need to move this term to the left hand side. So, we'll have a quadratic expression here and 0 on the right. After subtracting 2 from both sides, we want to try and factor this expression. We find factors of negative 10 that sum to positive 9. Those factors are negative 1 and positive 10. We use factoring by grouping to get these two factors equal to 0. Then, we can set each of these factors equal to 0 to solve for the values of u. U can equal negative 2, or u can equal 1 5th. So, since we know u equals negative 2 and u equals 1 5th, we can say b plus 3 equals negative 2 and b plus 3 equals 1 5th. We solve for this value of b to get b equals negative 5. Over here, we'll subtract 3 from both sides to get b equals 1 5th minus 3. We'll change 3 into 15 5ths and then subtract to get negative 14 5ths.

## u Substitution Practice 5

Alright, here's our fifth and last practice problem for this lesson. What do you think the solutions would be for x?

## u Substitution Practice 5

For this equation, x can equal negative 3 or 8 3rds. Good job if you got these two correct. I hope you're becoming a pro at u substitution. First, we'll perform the substitution and let u equal x minus 2. We'll replace x minus 2 with the variable u. Now that we have this equation, we need to move this 10 to the left-hand side. So, we'll have a quadratic equal to 0. We factored this quadratic expression by finding the factors of negative 30 that sum to positive and then use factoring by grouping to get u plus 5 times 3u minus 2 equal to 0. We use this factoring to solve for the variable u. So, we can have u equal negative 5, or u equals 2 3rds. Finally, we use those solutions of u to figure out the values of x. This was the original variable in our first equation. So, since u equals negative 5. x minus 2 equals negative 5. And since u equals 2 equation, so x can equal negative 3. Or x can equal 2 and 2 3rds. The mixed number, or 8 3rds, the improper fraction. Nice problem solving if you got here.

## u Substitution Challenge 1

Here's our first challenge problem for u-substitution. Notice that we have fractional exponents in our original equation. What type of substitution do you think we can use? And be careful when you go back to solve for the original value of a. When you think you've got it, write your answer here. Now of course, since this is a challenge problem, there won't be a solution. Try your best, and if you get stuck, head to the forums and discuss it with some other students.

## u Substitution Challenge 2

Here's our second challenge problem for U-substitution. Now again, there won't be a solution video for this problem, but there are two different ways that you can solve it. You can either use U-substitution by making some sort of substitution for these negative exponents and this variable. Or if you remember what a negative exponent means, you can rewrite this entire equation and solve it using something you've already done before. I hope you take on this challenge.