Algebra 1 (Eureka Math/EngageNY)
Sal finds the values of coefficients a and b that make (3x^a)(bx^4)=-24x^6 true for all x-values.
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- Aren't there infintly many solutions for a and b in this equation? All you have to do to get another correct solution if you choose an a other than 2 (2:32) is adjusting the value of b, right?(10 votes)
- You can't have multiple solutions for the equation in the video. There is only one solution for
b, but you can have different form for the solution.
ais only 2 for
a+4=6, however, you can write it as 4/2, 8/4 and so on.
bis only -8 for
3b= -24, however, you can do as I did with
Mathematics is so order.(10 votes)
- why did we divide 3 on both the sides?(2 votes)
- Look at the fractions. If I multiply b by 3 and then divide by 3, I canceled out my multiplication and am left with b. Plus, whatever operation I do to one side of the equation, I do to the other side. It all goes back to solving two-step equations that had a single variable.(4 votes)
- how do you reconcile if its a monomial or a polynomial(1 vote)
- A monomial is a polynomical with only one term (no add or subtract sings), A binomial has two terms (so you would see 1 addition or subtraction), a trinomial has three terms (so you would see 2 add/subtract signs), etc.(2 votes)
- How do we know what variables we are solving for? In this problem we had a, b, and x but only solved for a and b. Do we always only solve for 2 variables out of 3? Which 2 variables do I solve for?(1 vote)
- wouldn't the break down (from FOIL be) 3b+3x^4+x^ab+x^4? So breaking it down, 3 times b = 3b, 3 times x^4 = 3x^4, x^a times b = x^ab, x^a times x^4 = x^(a+4). How did the 3x^4 get combined with the x^(a+b)?(1 vote)
- You don't use FOIL with monomials. (3xᵃ)(bx⁴) gets multiplied like any two numbers, like (4)(x)... so it's 3xᵃbx⁴. Then you combine the x terms to get 3bxᵃ⁺⁴. What you're doing there would make sense if it was (3+xᵃ)(b+x⁴) instead.(1 vote)
- Ok, so I am going over this in Algebra 1 obviously but my question is that in my book it says apply power rule one and power rule two to this specific section . Is there a section in this course where sal explains it ?(1 vote)
- [Voiceover] We have three times X to the power of A times BX to the fourth power, and we have that equalling negative 24 X to the sixth. And so what I'd like you to do is pause this video and see if you can figure out what A and B need to be. Well to figure this out we can just multiply out this left-hand side, so let's do that. So what we have right over here, we can just rewrite that as three times X to the power of A times B times X to the fourth. And we can rewrite this as, we can just change the order, we're just multiplying four different things. This is the same thing as three times B times X to the A, times X to the fourth. And if you were doing this on your own, you wouldn't have to do all of these steps, but hopefully this makes it clear what's actually going on. And what's this going to be? Well, we have our, let me do this in a new color, we have our three B here. So that is three B, and what's X to the A times X to the fourth? X to the A times X to the fourth. Have the same base raised to different exponents and I'm multiplying the two. Well we know from our exponent properties, this is going to be the same thing as X to the A plus four. X to the A times X to the fourth is X to the A plus four, we're just adding the exponents because we have the same base and we're multiplying these two. So we now have that three B is equal to X to the A plus fourth power. And this is going to be equal to what we have on the right-hand side. So this is going to be equal to, put in that same color, negative 24 X to the sixth. So what could we do now? Well we can recognize that three B is going to have to be equal to negative 24 and the A plus four right over here is going to need to be equal to the six. So let's write that down. Three B is equal to negative 24. Three B is equal to negative 24, you might be able to do that in your head, or if you want to do it a little bit more systematically you could divide both sides by three to solve for your B. And you get B is equal to negative eight. And we can also say that A plus four is equal to six. Subtract four from both sides, you get A is equal to two. And we are done. And you can check it if you want. We can rewrite it as, if we say A is equal to two, we can say three X squared times, instead of B we know that's negative eight, negative eight X to the fourth. What is that going to be equal to? Three times negative eight is negative 24, X squared times X to the fourth is X to the sixth. And once again as I said, you wouldn't necessarily have to do all of these steps if you're doing it on paper, but this is what you're doing even if you look at this, if you looked at the original problem you'd say okay, the coefficients I can multiply those, three times B to get this coefficient over here. So three times B is negative 24, B would have to be negative eight. So B would have to be negative eight here. And you can say X to the A times X to the fourth is X to the sixth. So, well let's see, we're gonna exponents, A plus four is gonna B equal to six, A must be equal to two. So you might be able to do it that simply, but this is what you're doing, even if you're doing it in your head like that.