The easiest way to solve this problem is to factorize the variable part: a³-a²=18 a²(a-1)=18 Now we look and think, what number raised to the power of 2 times another number gives us 18? This lets us write the equation as: a²(a-1)=9×2 By comparison: a²=9 and a-1=2 a=±3 a=3 Since a has to be the same sign, we only have one answer a=3. (This step is faster for the brain to recognize than typing it.) Then you go on by solving the rest by long division and get the complex answers to the quadratic equation.

@anuragg7007

Жыл бұрын

Exact Same Solution / Procedure came in my mind 👍

@CalBruin

Жыл бұрын

Yeah, that is exactly how I solved. I thought there was something clever I missed or did not know.

@atif512

Жыл бұрын

Degree 3 polynomial, this method only yields 1

@mahmoudaboualfa5136

Жыл бұрын

@@atif512 read the last three lines

@heco.

Жыл бұрын

it also could be 6×3 so i don't know if this is the right way to do it

Right around here 1:50, I would recommend also bringing in the fact that it's impossible for any roots to be negative, as then the a^3, a^2, and the a^0 terms would all be negative, leaving no positives to balance it out to 0. Thus, no negative solutions exist, halving the guess and checking you'll need to do

@ingiford175

Жыл бұрын

Yep, and if you know more, you can look at the problem at 0:36 and tell that there is one positive root and no negative roots. Which limits the search case for your answer. By counting sign switches, and a quick substitution and check sign switches after substitution. +a^3-a^2-18=0 has one sign changes so it has 1 real positive solution insert -b for all the a's and see what signs are -b^3-b^2-18=0 has no sign changes so it has 0 real negative solutions. If you have a value that is 2 or more, your possible solutions are that, or that -2, until you get to 0 or 1. so if there are 5 sign changes in the test, you have either 5, 3, or 1 solution.

@mikefochtman7164

Жыл бұрын

Overall I agree with your conclusion. But how can you say that a^2 would be negative? Surely a negative root, if one existed, squared, would be positive?

@ingiford175

Жыл бұрын

@@mikefochtman7164 There is a - sign before the a^2 in the original equation, so that term is negative if a is positive (or negative due to square)

@ingiford175

Жыл бұрын

Look up Descartes Rule of Signs

@The_Scattered_Man

Жыл бұрын

Yeah... kind of feeling like the words "solve for all roots" and "solve over the complex numbers" COULD have been meaningfully included with the problem. (Hell, without ANY instructions, one COULD interpret it as "Solve for a matrix that satisfies these conditions" (assuming that "18" means "18 times the appropriate Identity Matrix"))... 🙂 [Sure, there's not reason to assume that, but there's no reason NOT To assume that either. ]

Observe that -18 can be rewritten as -27 + 9, s.t. difference of cubes and squares can be applied, we have a^3 - 27 - (a^2 - 9) = 0 (a - 3)(a^2 + 3a + 9) - (a - 3)(a + 3) = 0 (a - 3)(a^2 + 2a +6) = 0 Thus, a = 3 or a = -1+-isqrt(5) .

@extremelynoobgaming4742

Жыл бұрын

Wow it's sooo easy...

@flowingafterglow629

11 ай бұрын

Yes, that's a much better answer

The number in the equation can be rewritten as 27-9 =3^3-3^2. Now the equation can be written as: a^3-3^3-(a^2-3^2)=0 Now we have a difference of two cubes and difference of two squares. The problem now reduces to the same step after division.

@aguyontheinternet8436

Жыл бұрын

this is crazy insane witchcraft, how does one look at 18 and go _"ah, of course, a simple difference of cube and square of three"_

@IAM-lw4cj

Жыл бұрын

@@aguyontheinternet8436 Thank you

@theblinkingbrownie4654

Жыл бұрын

​@@aguyontheinternet8436idk about them but i turned a^3-a^2 into a^2(a-1) and knew 18=1*18=2*9=3*6, luckily 2*9 works as a^2(a-1)

@mike1024.

Жыл бұрын

​@@aguyontheinternet8436it's exactly what I spotted too lol. When you've done enough with numbers, you just see stuff like this. I was planning on using long division, but I agree that there is often a slick way to avoid long division and use factoring instead.

@cdmcfall

Жыл бұрын

Why even worry about differences of cubes and squares if you immediately recognize that a^3 - a^2 = 3^3 - 3^2? You've already come up with a solution for "a" that can then be used to factor the polynomial. I guess it's as easy as synthetic or even long division, but it seems like an easy way to mess up signs if you aren't careful: (a^3 - 3^3) - (a^2 - 3^2) = (a - 3)(a^2 + 3a + 9) - (a - 3)(a + 3) = (a - 3)((a^2 + 3a + 9) - (a + 3)) = (a - 3)(a^2 + 3a - a + 9 - 3) = (a - 3)(a^2 + 2a + 6) = 0

You can also use the Horner-Scheme to get the polynomial division. It‘s much faster and you just need the coefficients of your polynomial and fill in your guessed solution a=3.

Use synthetic division... You only use the constant, not variable. Make sure to use fillers, like +0a. Also bring the first constant down automatically. Find the first value of a, which we discovered is 3(a-3=0), put that in top left for reference. Multiply the top left number by the furthest right number at the bottom, place it below the next term, and add. Put the answer in the bottom and repeat. See here: 3 | 1 -1 0 -18 - V 3 -------- 1 2 3 | 1 -1 0 -18 - V 3 6 18 -------- 1 2 6 0 After finishing, put the variables back, but with 1 less power, because we divided by a(what it equals) 1a^2+2a+6 Then solve from there. It even matches his, but is easier.

@mittarimato8994

Жыл бұрын

isn't this exactly the same method as used in the video, but just different semantics?

@thetroiimaster

Жыл бұрын

@@mittarimato8994 no, what he did in the video is long division. And he used a variable in it. What this is isn't even division, it's multiplication and addition, and it uses not the variable, but what it equals. It is also a much easier method than that.

@mittarimato8994

Жыл бұрын

@@thetroiimaster looks exactly the same to me, except variable is 'hidden'

@thetroiimaster

Жыл бұрын

@@mittarimato8994 not the same, division and multiplication are not the same. His is longer and more complicated, while this is straightforward. Also, don't like your own comment, that's cringe.

@michaelgunn7609

Жыл бұрын

@@mittarimato8994 Though long division and synthetic division appear to be different, it can be shown that they produce the same result, i.e., quotient. . My students sometimes stumble over the details of long division, but they handle the less complex synthetic division almost without fail.

This somehow reminds me of the time when one of my friends asked the teacher if 0°C + 0°C = 64°F. The teacher was so confused she didn't come to school after that for three days in a row. Edit: I have summoned mathematicians and physicists in the comments section.

@jeffmartin5419

Жыл бұрын

Wouldn't it be 523.67 F? (because it's 273.15 K twice)

@itsanandaman

Жыл бұрын

It will be 32f+32f=64f

@mohammedelmohandes7629

Жыл бұрын

Dood it’s just 32 as there is no temperature added thus it’s only 32 F

@mxstila6220

Жыл бұрын

*Warning: This comment is super long and is my sad attempt to show my solution to this problem, which is 32ºF. I don’t know if this makes any sense or if anything is correct, but if you dare to read it, you can click read more I guess. Also uh I didnt watch the video so hopefully the things I am saying weren’t disproved in the video or anything; I am an incoming sophomore this year so please forgive me lol* So how I explain it is the system for Fahrenheit starts at 32 degrees rather than 0 degrees. I think I wrote in a comment that any given number can be represented as _z + v = n_ where _z_ is the “zero” value of the system (which for this purpose is the temperature of the freezing point), _v_ is the value of the number relative to the zero (so the number minus the _z_ value) and the _n_ is the actual number that we write.* This usually isn’t necessary to think about, as the zero of most systems is literally *0* (hence the name “zero” that I made up), and even for silly systems like Fahrenheit temperatures, we just add/ subtract 32 to find the _v_ in the formula. However, when more than one value is represented, the _z_ value still should only be added once. This is because you add the _values _ relative to the zero] together and not the actual numbers together, and then add the zero. You can think of subtracting the values that are below freezing from the temperatures. For example, 50ºF has 32º worth of values that are below freezing, so the value would actually be 50º - 32º, or 18º. As stated before, adding the zero to the value makes the actual number. When adding temperatures together, you can add each number, then subtract the 32º below freezing from each number (for example: 50 + 50 has two numbers with 32º below freezing, so you would subtract 32º twice, which would ultimately mean subtracting 64º) to find the temperature _value_ , then add the zero, 32, to that value find the actual number. If you did this, the equation _0ºC + 0ºC = x_ would look like: 0ºC = 32ºF (convert to Fahrenheit) n = 32 z = 32 v = ? 32 + v = 32 (finding the value) -32 -32 v = 0 0 + 0 + 32 = x (adding the two values and the zero together) x = 32 Alternatively, it could look like: 0ºC = 32ºF Of those 32ºF, 32º are below freezing, so we would subtract 32 from each number to find the value Option 1) 32 - 32 + 32 - 32 = v v = 0 Option 2) v = 32 + 32 - 32 - 32 32 + 32 = 64 64 - 32 - 32 = 0 From that value, the zero should be added to convert it to the actual number, so: 0 + 32 = 32 x = 32 Or you can just do the equation in celsius, where the _z_ value is actually *0* . Hopefully this made sense! I never have done anything like this, so if you find any errors or have any questions, please ask me. I don’t really think this has any practical use, but it does solve the problem I guess lol *The equation can alternatively be written as _n - z = v_ Edit: I DIDNT NOTICE HOW LONG THIS WAS WHAT THE HECK

@alexandra_reznikov

Жыл бұрын

@@mxstila6220 My teacher had explained somewhat the same thing if I remember correctly but then she had forbidden us to talk about equations in class. Thanks for taking out your time to write this!

Most Algebra textbooks will introduce synthetic division with the rational root theorem. Direct substitution, as you’ve demonstrated, is a little quicker. However, it’s nice to show your students the technique of synthetic division. Some students will gravitate to synthetic division over the long division technique and some will not. Nice video 😉👍

@itzmrinyy7484

11 ай бұрын

Synthetic division makes things SO much easier, love solving these cubics

@BoxStudioExecutive

7 ай бұрын

Funnily enough, I was on the math team in high school, did AP Calc BC and NEVER learned polynomial/synthetic division.

@eatingyoshi4403

7 ай бұрын

@@BoxStudioExecutive I did it in algebra 2 and for like a couple days in precalc, its possible that you just forgot about it or the teacher was running out of time.

That how SAT made simple looking equation to a compicated equation.

@epikherolol8189

Жыл бұрын

Complicated?? Bro i can see in 5s that a=3 is a solution. So dividing the equation by (a-3) will give quadratic then we use factoring/formula

@aparnarai3708

Жыл бұрын

Well all you need to know is how to take common from a³ - a² and then it's pretty simple. But the main reason this video is here is to make the concept clear It's 3 that's why we found it easily What if it's something like A¹⁶ - A¹⁵ = 122070312500

@Mrgodclipz

Жыл бұрын

I love your profile picture

@zhenningli2965

Жыл бұрын

@@aparnarai3708Then we need the graphic calculator😊

@Allangulon

11 ай бұрын

3x3x3=27. 3x3=9. 27-9=18. a=3.

I am happy you discussed the general case also instead of focusing only on the current question

When looking for whole number solutions, the form a^2*(a-1)=2*3*3 pretty quickly yields a=3, but the number of things to check grows for more highly composite numbers.

At least for this example, you don't need to do long division once you have (a-3), you can simply compare coefficients. You can see from the fact its a cubic equation with coefficient of 1 for a^3, and by the fact that the constant term is 18, that it will be (a-3)(a^2 +ka + 6) Then simply compare coefficients and k must be 2. Also alternatively to using the quadratic formula you can complete the square: (a+1)^2 = -5 a+1 = +/- iroot5 a = -1 +/- iroot5

I do love this theorem... The first exam I won at University was about this topic... BUT when you can just guess and try small numbers into the equation, and it works, you can't tell with grandiosity it is a "Math Olimpiad Problem"

@carultch

Жыл бұрын

He probably simplified it from the Math Olympiad problem it was based on.

You can use Ruffini as well, instead of the standard polynomial division

instead of long division, you could instead use synthetic division to make it much faster.

@gamingofrishit888

Жыл бұрын

What is that?

@tekgewet

Жыл бұрын

@@gamingofrishit888 you use the coefficients of each term in the things you’re dividing and the h part in a-h (thing you’re dividing by) and set up all the coefficients in a line. then, you drop the coefficient and then multiply by the h and move it to the next coefficient and add. repeat this until you have all of them, and those are coefficients to the equation, except for the last one which is the i part iirc (this is a horrible explanation just google it)

@chloeuntrau4588

Жыл бұрын

yes...I had it in a few seconds...and I am zero in math!

@lavrentizapadni747

Жыл бұрын

@@tekgewet A valiant attempt nevertheless - good man!! 👍👍👍

@TypoKnig

Жыл бұрын

Synthetic division is not commonly taught, which means the video would have to take time to explain it. I think this awesome technique should be more commonly used.

Very clear! Thank you!

3: 27 (which is 3 cubed) - 9 (which is 3 squared) = 18 I truly didn't even have to think about it. It just jumped out. Thank you for your efforts. May you and yours stay well and prosper.

It is interesting to see that the difference between a² and a³ for 2 is 2*1*2, for 3 it is 3*2*3, for 4 it is 4*3*4, and so on. So it always is a²*(a-1).

@killianobrien2007

Жыл бұрын

Factorisation

Thank you for your sharing

I hadn't thought about factoring out a-3 from the original polynomial after guessing the first root. Nice.

After looking at the coeffients, I also look and compare a^3 and 18. What factors of 18 are great enough that I can subtract 18 from a^3 and not go negative?

Nice question Thank you

You can simplyfy the first part a lot. Just factor twice variable a out of the left part of the original equation. That gives a*a*(a-1)=18 If you then factorize 18 you get a*a*(a-1)=3*3*2 So a=3

I did it by hit and trial. a³-a²=18 (Given) Trial 01: When a=2, then a³-a²=(2)³-(2)²=8-4=4≠18 Trail 02: When a=3, then a³-a²=(3)³-(3)²=27-9=18 Thus, a=3 satisfies the given equation.

This method reminds me of a rule i have been using since 2nd year of high school, called the Ruffini Rule. It works really similar to the method used here, but for less experienced people it may seem more viable in case they get messed up. It works like this: After finding that 3 is an answer to the method, we put it in a table and rewrite the main equation, leaving the known term alone. Doing a set of calculations you end up with 0 at the end, and what you've written during those calculations are the coefficients to the equation you need to multiply the first term (a-3) with.

Thanks from Morocco

My trick for these sorts of problems, when I see them in youtube thumbnails, is guessing. Due to the size of the number 18, I thought a = 3 might be a nice first guess, as these problems often have integer solutions just by how they are made to be. 3³-3² = 27-9=18 Usually it doesnt work very first try but it works well for these artificial problems.

Trial and error works. a=2 --> a^3 is too small. a=3 --> yep, that's the solution. That's a simple trick! 🙂 Also, a^2 is a factor of the l.h.s. So for an integer solution, we have to find a perfect square that is a factor of 18. There is only one a^2 = 9, thus a=3. For additional roots, factor out (a-3) or use synthetic division, and solve the quadratic.

I got the same answers using pretty much the same method! But with extra (unnecessary) steps by substituting a -1 = u and solving for u from there. It's been a few years since I've done a math exercise like this. I feel both smart and silly for solving this using such an inefficient way. XD

Long division is very long method we used to learn in 5th grade Alternatively use diagnolly multiplying the first root with coefficients of other and get other roots..or synthetic division as well.

You're simply the Best

can you do a video on math professor Lou Kauffman's iterant origin of the imaginary number due to an asymmetric time shift as noncommutativity? Thanks

Its also possible to try and find a factor by adding and subtracting some number: a^3 - a^2 - 18 = 0 a^3 - 3a^2 + 2a^2 -18 = 0 Then you can isolate (x - 3) and factors it out so u get (a - 3)(a^2 + 2a +6) = 0 Though that is something thats not so easy to see.

It's much simpler to use the straight and simple Ruffini' s rule, once you recognize that a = 3 is a root, so it only takes a few seconds to solve it

Thanks for acknowledging Brahmagupta, and pronouncing it correctly! 👏👍

@carultch

Жыл бұрын

I'm getting mixed messages when I read other comments who claim Shridharacharya should be the namesake of the quadratic formula. What's the story behind whether Brahmagupta or Shridharacharya should be its namesake?

In the UK we call thst factor theorem, it's a fairly standard thing taught at A level maths for solving cubic equations. We sometimes get questions on it and it is useful to know but having a standard method to solve any cubic would admittedly be nice XD

a^3-a^2=18 Rewrite 18 as 27-9 (which is a cube and a square): a^3-a^2=27-9 Rearrange: a^3-27=a^2-9 Difference of cubes and difference of squares formula: (a^2+3a+9)(a-3)=(a-3)(a+3) We find the solution a=3 If a is not 3 then we can remove (a-3) from both sides of the equation: a^2+3a+9=a+3 a^2+2a+6=0 (-2+-sqrt(4-24))/2=-1+-i*sqrt(5)

Not sure why this came to mind, but I found the first root via: a^3 - a^2 = 18 a^2(a-1) = 3^2(2) a^2(a-1) = 3^2(3-1) a = 3 since it's the only value that fits the last iteration

👏👏👏👏👏👏👏👏👏👏 thank you.

Should it be easier if we write a^3 - 27 - a^2 + 9 = 0 then perform a long division?

I prefer using synthetic division rather than long division. While long division pretty cumbersome and time-consuming to execute, synthetic division only works around the coefficients, which is much faster.

I dont know but at first glance I know its 3 already. I didn't consider the 2nd solution but the way it solve is satisfying.

a^3 - a^2 = 18 => a^2(a-1) = (3(sqrt(2)))^2 => square rooting on both the sides a(a-1)^1/2 = 3 sqrt2 On comparing a=3. I think it's the easiest way to solve these type of questions.

The best way is after you use rational root theorem put what x equals say x=2 write down each coefficient for all pro-numerals including the constant then leave a space for a line of numbers below then draw a line for the first number write it below the line then above the line under the second number give it is x^3 1x2 write 2 then say it is 2x^2 2+2 write that below the line as 4 times 4x2 write 8 above the line under the third term and so on you know you have the answer if you end up with say the constant is 18 and the number below is -18 it will equal 0 and you will have a quadratic equation and just solve for x after tht

I left the 18 of the RHS and factored the LHS as a^2*(a-1). "Assuming" (since you did also) that the solution is integer then we're looking for factors of 18 like a^2 and a-1 and pretty quick a = 3 gives 9*2

Much simpler: a^2(a - 1)= 18. The only perfect square factor of 18 is 9. Therefore, a^2 = 9; a=3. 3^2(3-1)=18. Alternatively, use synthetic division.

Before watchng the video: a^3 - a^2 = 18 First, this is a cubic, so we want to find one root first, so we can simplify it to a quadratic. We will try to find a rational root with the rational root theorem... Except we are going to notice something extra. First, the classic RRT tells us that if there exists a rational root, then it is an integer (specifically a divisor of 18, or the negative of it) Suppose that n is a soluton. Then n^3 - n^2 = 18. Note that n^2 is a common factor on the LHS. So the right hand side is also divisible by n^2. This means that n=-3, -1, 1 or 3. Furthermore, note that if n is negative, then n^3 will be negative and n^2 will be positive. A negative minus a positive cannot equal 18. So, the only options are n=1 and n=3. Testing those two, we find that n=1 does not work, but n=3 does. This means we have a-3 as a solution. Next off, divide our cubic by a-3 to get rid of that root. (a^3 - a^2 - 18)/(a - 3) = a^2 + 2a + 6 This means that, if there are any other solutions, they will be found from the quadratic a^2 + 2a + 6 = 0 a^2 + 2a + 6 = 0 a^2 + 2a + 1 = -5 a+1 = +/- sqrt(5) i a = -1 +/- sqrt(5) i So, the solutions are a=3, a=-1 + sqrt(5) i and a = -1 - sqrt(5) i

@lol1991

Жыл бұрын

Now that has to be the easiest solution I’ve seen😮

The way I did it was rewriting the right handside of the equation as 27 - 9. a³ - a² = 27 - 9 Then adding/subtracting 27/9 from both sides. a³ - a² - 27 + 9 = 0 Rearranging the terms. a³ - 27 - a² + 9 = 0 27 is 3³. a³ - 3³ - a² + 9 = 0 Factoring out a "-1" from (-a² + 9). (a³ - 3³) - (a² - 9) = 0 Factorizing the difference of cubes subtracted by difference of squares. (a - 3)(a² + 3² + 3a) - (a - 3)(a + 3) = 0 Factoring out the common factor (a - 3). (a - 3)[(a² + 3² + 3a) - (a + 3)] = 0 Simplifying. (a - 3)(a² + 3² + 3a - a - 3) = 0 (a - 3)(a² + 2a + 6) = 0 a - 3 = 0 V a² + 2a + 6 = 0 a = 3 V a = -1 ±√5i

what i did was simply factor the lhs into a²(a-1) after looking rhs it was obvious 18 = 9 x 2 or a = 3. For bigger constant it might be hard to guess, but also the constant may have a lot of factors so it would be hard with the trick anyways. Cool theorem though.

I am confused. It seems to me that when you did polynomial long division you were dividing by zero (a - 3, with a = 3, should yield 0), which is undefined? Please explain.

This one I got very fast. I knew "a" had to be a very low number because the gap between the cube and the square will keep getting exponentially larger very fast. I tried 2 for "a" which was too low and then tried 3, and that's when I figured out it was 3.

I actually solved this in a different way. I'm pretty sure others did the same method as well, but I would very much like to share mine: Simplify the LHS a² ( a - 1 ) = 18 Squareroot both sides a [ sqrt( a - 1 ) ] = 3 sqrt( 2 ) Then a = 3 a - 1 = 2 Both equations prove a = 3

I think you should use synthetic division

It is pretty easy to see a=3 is a solution then continue by synthetic division and quadratic equation to find complex solutions. Another method I saw quickly was by adding 9 to both sides we could find difference of squares and cubes. a³ - a² +9 = 27 where 3² = 9 and 3³= 27 So a³-3³ = a²-3² --> (a-3)(a²+3a+9) = (a+3)(a-3) Thus a=3 is a solution. If a!= 3 then we can divide out (a-3) on both sides leaving a²+2a+6=0. Proceed with quadratic eqn to find complex roots.

i always appreciate how presh calls it brahmagupta's formula [:

Doing something like this in School, we’re doing Factorising trinomals and the process is very similar

Well, if you only want real solutions, then you can do this: a^3-a^2=18 a^2(a-1)=9*2 a^2=9 OR a-1=2 a=3 or -3 a=3 And then you can proceed with rational root theorem for imaginary solutions.

This reminds me of the integral domain Z{sqrt(-5)), in which 6 does not have unique factorization as 6 = 2 * 3 (solutions of x^2-5x+6) and 6 = (-1+sqrt(-5))(-1+sqrt(-5)) (solutions of x^2+2x+6). The two equations differ in the x term, with coefficient of -5 in the first case and +2 in the second.

Nice

Can u make a video how to we make this type video

Awesome explanation😊

the way i solve this question is by first i imagine what cubic numbers are when ± a quadratic number gives 18. turned out the perfect combination is 27-9 which is 3³ and 3².after that i form an equation a³-a²=3³-3²(from this we can see that the first solutions are 3) .then i changed it to standard form a³-3³-a²+3²=0. after that, i factorize the equation which then i get (x-3)(x²+3x+9)+(3-x)(3+x)=0.then i change the (x-3) to -(-x+3) which helps me to factorize the equation once again and form -(-x+3)[(x²+3x+9)+(3+x)]=0. next i compared the equation=0 and found the solutions as x=3 and x=-1±i√5.

Once you arrive at a=3 why isn’t that good enough ? Is the actual answer a= +or - 3 because both seems to work, -3 and +3.

If I remember correctly this was the default method we used in grade 9 to solve cubic equations

The value of a= 3 a³ -- a² = 18 can be written as a²(a -- 1) = 18 Write 18 as (9×2). a²=9 ; (a-1)=2 [Simultaneously] Therefore, a= 3

It is natural to solve easy problems correctly, but it is also important to solve them quickly. Since this problem is easy, it should be faster to find the term (a^3-a^2-18) by multiplication rather than dividing the term (a-3). In other words, from (a-3)(a^2+……), to change (-3a^2) to (-a^2), (……) should be replaced with (+2a+……). Next, from (a-3)(a^2+2a+……), to change (-6a) to (0), (……) should be replaced by (+6). So (a-3)(a^2+2a+6), the last (-18) matches (-3)×(+6)=(-18). Also, using another formula for the solution of the quadratic equation, if ax^2+bx+c=0(a≠0), b is even number (b=2b'), x=[-b'± √{(b')^2-ac}]/a, then x={-1±√(1^2-1×6)}/1=-1±i√5, it is easier and faster calculated.

2:20 "one only has one factor, which is one." yeah, sounds like I might be able to follow along 👌🤣

Well done

Rational Root theorem is very useful. It was taught in my school in 9th grade in Delhi, India here.

@martinkuliza

Жыл бұрын

we do know that Delhi is in india FYI

@sparshsharma5270

Жыл бұрын

@@martinkuliza There's a Delhi in US, Canada, Australia, China as well!

@martinkuliza

Жыл бұрын

@@sparshsharma5270 Well i live in Australia there's no Delhi here as for Canada and China that's not common knowledge Common Knowledge is that Delhi is in India. if you were in Canada and said in Delhi, Canada here i would let it go Example, here in Sydney we have TORONTO, but that also wouldn't be common knowledge, it would be common to assume United states so yeah, we know Delhi is in india, there was not need to say that , Unless you were in one of the others not commonly known

@sparshsharma5270

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@@martinkuliza I said so because when I once said Delhi, suddenly many others commented 'where' as if they never heard of such a place and started making assumptions with Internet then I had to specify. On a channel like MYD, there are tons of viewers globally so I think it's better to specify else someone would have definitely picked on me.

I was fondly hoping that you would use synthetic division...

We used something called Ruffini rule and it's basically the same but much faster. No need to do polynomial divisions

What about ruffini's rule? It's the same but easier.

Also applying Ruffini's method

I approached it by calculating the lowest integer value of a^3 that was greater than 18. It only took about 3 seconds to solve it that way.

Move 18 to the other side. We have a^3-a^2-18=0. By inspection, 3 is a root of the equation, therefore a-3 is a factor of a^3-a^2-18 (if a is a root of a polynomial, x-a is a factor). Using synthetic division/long division, we get the factored form of a^3-a^2-18 -> (a-3)(a^2+2a+6)=0, leading us to the solutions a=3, -1+isqrt(5), -1-isqrt(5).

Best way is to find a simple solution by inspection, clearly 3 ==> a-3 is a factor of a^3 - a^2 - 18 Then seek factorisation: (a-3)(pa^2 + qa + r) Clearly, p = 1 (equate a^3) r = 6 (equate const) and q - 3p = -1 ==> q = 2 (equate a^2 term) So we get: a^3 - a^2 - 18 = (a-3)(a^2 + 2a + 6) Solve quadratic by completing the square ==> (a+1)^2 -1 +6 = 0 ==> a = -1 +_ sqrt(5) i (a = 3 also from linear factor) There are the 3 solutions

Finding factors by eye is faster, then do the stuff for the complex solution.

Is the rational root theorem really a trick?

tbh I just guessed some integer small solutions cus a^3 grows too faster than a^2 for a to be able to be a very big number and found the 3 root, then divided the polynomials to find the other ones

a =3 a^3- a^2 =18 a*a*(a-1)=18 Since a > 0, then a*a*a > (a-1)(a-1)(a-1) and (a-1)(a-1)(a-1) 18 and the nearest cube that is (a-1)^3 =8 a^3 = 27 ; hence a =3 and (a-1) =2 Hence a*a*(a-1) = 3*3*(3-1)= 3*3*2 = 18

In order for one of the roots to be 1, doesn't the sum of the coefficients need to be 1?

So what's i equal to?

Hey, I remember when I used to love maths

Solving a simple cubic equation: a^3 - a^2 = 18; a = ? First method: a^3 - a^2 = 18 = (9)(2) = (9)(3 - 10) = 27 - 9 = 3^3 - 3^2; a = 3 Missing two complex roots Second method: a^3 - a^2 - 18 = 0, (a^3 - 27) - (a^2 - 9) = 0 (a - 3)(a^2 + 3a + 9) - (a - 3)(a + 3) = (a - 3)(a^2 + 3a + 9 - a - 3) = 0 (a - 3)(a^2 + 2a + 6) = 0, a - 3 = 0 or a^2 + 2a + 6 = 0, (a + 1)^2 = - 5 a = 3 or a = - 1 ± i√5 Answer check: a = 3, a^3 - a^2 = 18; Confirmed in First method a = - 1 ± i√5; a^2 + 2a + 6 = 0, a^2 + 2a = - 6, a^2 = - 2(a + 3) a^3 - a^2 = (a^2)(a - 1) = - 2(a + 3)(a - 1) = - 2(a^2 + 2a - 3) = - 2(- 6 - 3) = 18; Confirmed Final answer: a = 3, a = - 1 + i√5 or a = - 1 - i√5

How do you make this type of videos. Can you please tell me.??? I want to make KZread videos, regarding mathematics. Please help ....

There is also the generic formula for cubic equations, but that would be overkill for this problem.

We can also use the sum of roots sum of product of roots and products of roots formulae

Love from India🇮🇳

Very nice, but is there a similar solution, that works for rational numbers?

@carultch

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Yes. Consider your polynomial in general to have the form: A*x^n + B*x^(n-1) + other terms ... + K = 0. All of your rational candidates for solutions will be related to the coefficients of A and K. The factors of K are the numerators, and the factors of A are the denominators. You then form fractions with this information, and use trial and error. As an example, consider 2*x^3 + x^2 - 8*x + 3 = 0. Our K-constant is 3, and our A-coefficient is 2. This means that factors of 2 / factors of 3, will be the possible rational solutions. Both of these are prime numbers, so that limits our search. The possible solutions are -3, -3/2, -1, -1/2, and +1/2, +1, +3/2, and +3. Trying x = +3/2, we see that it is a solution to this equation. The other two solutions are x = 1 +/- sqrt(2).

@firiasu

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@@carultch Cool! But for irrational numbers there no simple solution probably.

I prefer using the remainder theorem, where you skip one line, go to the third line and take a-3 as a factor three times, and calculate the second line with the help of the third line

Me: Mission accomplished at 3:00! You: Too simple. Let's juggle some more balls...

Somehow somewhere when you started the division after a=3, I lost the connection.

Rewriting the original expression by expanding both sides: a * a * (a-1) = 3 * 3 * 2 ...well, that was easy. Prime factorization is not only useful in The Cube, it seems.

I think the "trick" they were saying would make the solution easier to find is to recognize that a^3 - a^2 = (a-1)*a^2, and 18 = 2*3^2 = (3-1)*3^2. If all you need are real roots, then you are done, because a^3 - a^2 - 18 = (a - 3)*(a^2 + 2*a + 6), and the discriminant of a^2 + 2*a + 6 is -20.

Works only if there is an integer solution. The integer solution could also be found like this: right side) 18 = 3 * 3 " 2 left side) a^3 - a^2 = a * a * (a -1)

a^3-a^2=a^2(a-1)=18=3*3*2=3^2(3-2); So one root is a=3. Extract the factor from the expression and then solve the quadratic.

Or you could let a^2 = y so that y^2-y-18=0 and solve using quad formula and then resubstitute

This is what I did, except that I just saw instantly that 3 was a solution and knew I had to divide by a-3.

Instead of long division method we can use synthetic division method to solve the problem in short time

My attempt I don’t know what trick he will refer to, but I’d like to use the time-honored trick of hoping that there’s an integer solution and factoring stuff. Note: If this didn’t work, my next approach would be using the rational roots theorem Edit: lol that was the trick a^3 - a^2 = a^2(a-1) 18 = 2 * 3 * 3 The only square that divides 18 is clearly 9, so a = 3 or a = -3 are the only possible integer solutions. 3^2(3-1) = 18 (-3)^2(-3-1) ≠ 18 The only integer solution is a=3 Now rearranging and dividing by a-3, we obtain the following assuming a≠3 a^2 + 2a + 6 = 0 This has no real solution, but if you don’t care, then we have more work to do. a = (-2 +- sqrt(4 - 24))/2 a = (-2 +- sqrt(-20))/2 a = (-2 +- 2i*sqrt(5))/2 a = -1 +- i*sqrt(5)/2