Solving a simple cubic equation. A trick you should know!
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- Опубліковано 21 вер 2024
- This is a popular question on UA-cam and someone requested that I present a solution. There is an important mathematical theorem you should know!
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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.
Exact Same Solution / Procedure came in my mind 👍
Yeah, that is exactly how I solved. I thought there was something clever I missed or did not know.
Degree 3 polynomial, this method only yields 1
@@atif512 read the last three lines
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
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.
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?
@@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)
Look up Descartes Rule of Signs
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) .
Wow it's sooo easy...
Yes, that's a much better answer
That’s how I did it, but this was a great lesson on application of the rational root theorem.
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.
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"_
@@aguyontheinternet8436 Thank you
@@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)
@@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.
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
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.
isn't this exactly the same method as used in the video, but just different semantics?
@@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.
@@thetroiimaster looks exactly the same to me, except variable is 'hidden'
@@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.
@@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.
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 😉👍
Synthetic division makes things SO much easier, love solving these cubics
Funnily enough, I was on the math team in high school, did AP Calc BC and NEVER learned polynomial/synthetic division.
@@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.
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.
Wouldn't it be 523.67 F? (because it's 273.15 K twice)
It will be 32f+32f=64f
Dood it’s just 32 as there is no temperature added thus it’s only 32 F
*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
@@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!
That how SAT made simple looking equation to a compicated equation.
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
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
I love your profile picture
@@aparnarai3708Then we need the graphic calculator😊
3x3x3=27. 3x3=9. 27-9=18. a=3.
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.
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.
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.
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"
He probably simplified it from the Math Olympiad problem it was based on.
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.
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.
I am happy you discussed the general case also instead of focusing only on the current question
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
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)
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.
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.
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.
instead of long division, you could instead use synthetic division to make it much faster.
What is that?
@@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)
yes...I had it in a few seconds...and I am zero in math!
@@tekgewet A valiant attempt nevertheless - good man!! 👍👍👍
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.
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).
Factorisation
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
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
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.
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
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
Now that has to be the easiest solution I’ve seen😮
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
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.
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
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 . Hence find the nearest cube that is > 18 and the nearest cube
that is < than 18 . 8 and 27 or 2^3 and 3^3
(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
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.
You can use Ruffini as well, instead of the standard polynomial division
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
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
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).
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.
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
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
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.
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.
Can't just we do it like this:
a³-a²=18 so,a²(a-1)=18 we can say that a² is a perfect square number and 18 can be written as 9×2 so therefore a²=9 so a=3.It also satisfy that a-1=2.It would be easy in a competitive exam.
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.
This is the standard procedure...
Do not call it " a trick "
Your videos are great...
Thank U
I admit I too was hoping for some clever way to look at this other than literally the stuff that is taught in an Algebra II class.
It's a trick.
Thanks for acknowledging Brahmagupta, and pronouncing it correctly! 👏👍
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?
I didn’t use math, I used logic. I figured it like this:
18’s a pretty small number, so a will most likely be a one-digit number.
18 is a multiple of 9, but 9 would be too big of a number to be a if you’re going to cube and square it, so…
9 is a multiple of 3, and…
3^3 = 27
3^2 = 9
27 - 9 = 18
(Did it a lot quicker in my head)
a³ - a² = (a -1) a² = 18. If a is an integer, than a-1 and a are consecutive integers, one is odd and the other is even. If a were even, then a² is dividable by 4, but 18 isn't. So a-1 has to be even, and a² an odd square number. OTH 18 factors into 18 = 2 * 3 * 3, so a - 1 = 2 and a =3.
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)
Solving for integer values of a. a^3-a^2 = 18. Factor to a^2(a-1) = 18. So a^2 must be a factor. 9 is the only square that is a factor of a. a = 3 and a-1 = 3-1 = 2 is also a factor. so the answer is 3 as (3x3(3-1) = 18.
I felt very intelligent now because I just thought "ok, something close to 18 that is a cubic number. 2 is 8 and 4 is 64, so it's 3. 3 cubed is 27, 27 - 18 is 9, 9 is 3 squared. solved"
a³-a²=18
a²(a-1)=18
a*a*(a-1)=2*3*3 (by prime factoring)
You can immediately see a=3 is a solution.
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
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 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.
The rational root theorem works just fine when at least one of the roots is rational. However, that is an artificial problem, made up by a math teacher to be solvable.
Something more interesting would be to provide a solution for a^3 - a^2 = C, where C is any value. If C = +12, the rational root theorem doesn't do any good.
We used something called Ruffini rule and it's basically the same but much faster. No need to do polynomial divisions
If I remember correctly this was the default method we used in grade 9 to solve cubic equations
2:20 "one only has one factor, which is one."
yeah, sounds like I might be able to follow along 👌🤣
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
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.
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
l'équation s'écrit a^3 - a² -18 = 0 nous apercevons une racine évidente a = 3
nous pouvons factoriser (a - 3)(a² + 2a + 6) = 0
soit donc une racine réelle x = 3 et deux racines complexes conjuguées x1 = -1 +irac(5) et x2 = - 1 - irac(5)
Should it be easier if we write a^3 - 27 - a^2 + 9 = 0 then perform a long division?
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.
I simply factorized a³ - a² and got a²(a - 1) = 18.
Now this means a number squared times one less than that number is 18.
I know 9 × 2 = 18
3² (3 - 1) = 18
Therefore a = 3.
The working looks long but it is done mentally in seconds.
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 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.
3:17 instead of long division we can use synthetic division
It will be easier :)
"Easier" or "more easy" but not "more easier"
@@l.h.308 I think srinivasRamanujan123 was trying to be more funnier... 😛
@@l.h.308 thank you for correction
Yes, he was most funniest! @@timeonly1401
Doing something like this in School, we’re doing Factorising trinomals and the process is very similar
Super simple trick.cosider a^3 - a^2= a^2(a-1)=9 × 2 = 3^2 × 2 = 3^2 × (3-1) .then you can see that a = 3
Me: Mission accomplished at 3:00!
You: Too simple. Let's juggle some more balls...
write 18 as prime factors, rearrange equation given till fits, done a=3
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.
I should know but I don't 😁
...yet I do know a few "hack arounds":
a²(a-1) = 3² * (3-1)
So it's obvious that one of the roots is 3, knowing that I could try using "long division" and divide the entire expression by (a-3) this would give me a quadratic equation solving which I'd find all the roits - if more of them exist
I solved it like this:
a^3 - a^2 = 18
a^2(a - 1) = 2 * 3^2
a^2(a - 1) = 3^2(3 - 1)
so a is clearly 3
but then you dont get a = -1 +- i * sqrt(5)
@@zectus i actually dont care about imaginary solutions I think they are silly and obsolete
1 only has 1 factor which is 1. FINALLY SOMETHING I KNOW ABOUT!
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.
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.
There's actually a much simpler solution to the problem. If a^3 - a^2 = 18. Then it simply means, a*a*(a-1) = 18. Well to get 18, there are limited set of possibilities. Product of 18,1; 9,2; 6,3; and the product of the negatives -18,-1; -9,-2; -6,-3. If we look at say 9,2 it is simply 3,3,2 which fits the description of a*a*(a-1). The rest of them don't.
Worth noting that if there is a rational root to the cubic equation, it will be one of the fractions found with this method.
However, if you assign random integer numbers to the coefficient, most cubic polynomials will not have any rational solutions.
Of course in general, when teaching this subject, teachers will give cases with at least one rational root so you can reduce it to a quadratic equation like done here.
But in one case (I think that do to a mistake by the teacher) I spent HOURS trying dozens of fractions and NONE was a root. It was case where both the coefficient of X^3 and the independent term had lots of factors. Like 12 and 20. +/- (1,2,3,4,6)/(1,2,4,5,10,20) is 40 fractions to try by brute force, only to find none that works!!!!
Do you remember if you found any cases where the value of the polynomial changed sign from one trial root to the next? That would give you and interval where a solution exists. Then you could use numerical techniques to find a root - assuming you were taught numerical techniques in that class.
@@TypoKnig ... That was decades ago, I don't remember. But I am quite sure I did not put all the candidate fractions ordered along the real numbers line and tried them in that order. So nope. And if I had done that, I might have realized that there has to be a root between these 2 candidates, but I didn't know numerical methods yet as to try to approximate a solution. But I could have put the formula in Excel or calculator and approximate manually using the half-point method, which I find faster than any numerical method (not necessarily faster-converging, but definitively faster to implement and get a satisfactory approximation).
When I was teaching this, I sometimes used to set irreducible cubics (ones with no rational factors) to emphasise the importance of the factor theorem/rational root theorem shown in this video. It means that you don't keep searching endlessly for roots by trial and error, there will be a finite set of candidates, and once the student has eliminated them all, they know there are no factors. Another good teaching ploy is to set a cubic which is in fact a perfect cube. This highlights the value of extracting a factor once it has been identified, as those who continue with the trial and error method are looking for further roots, when they already have the only one.
There is also the generic formula for cubic equations, but that would be overkill for this problem.
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.
Or you could let a^2 = y so that y^2-y-18=0 and solve using quad formula and then resubstitute
Love from India🇮🇳
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
Also applying Ruffini's method
I found the first root by noticing that a^3 - a^2 = a^2 ( a - 1 ), or a x a x (a-1). Now we see that 2 x 2 x 1 is too small 4 x 4 x 3 is too big, but 3 x 3 x 2 works!
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.
a³-a²=18 -> a*a*(a-1)=18
looking for integer solutions: factors of 18: 1,2,3,6,9,18, and we're looking for consecutive pair.
1 and 2: 2*2*1=4, nope - also, doesn't have a multiple of 3 in it...
2 and 3: 3*3*2=18, yep
I was fondly hoping that you would use synthetic division...
Finding factors by eye is faster, then do the stuff for the complex solution.
In 4th grade elementary school, we had to memorize square numbers from 1 to 25 and cube numbers to 15.
I don't remember them all, but about 5 seconds after seeing the tumbnail of that video, I knew that 27 (3^3) minus 9 (3^2) is 18.
I was thinking ov3r different possibilities for a in the beginning to see if I could come up with an equation without help. I tried plugging in 2, so '8-4', and then plugging in 3, so '27-9' to see if there was a visible relation. Then I realized that 27 - 9 IS 18. So a = 3.
Haven't come up with an equation yet, I'm about to watch the video.
For some reason skipped that 18 is a factor of 9. So the roots must either both be factors of 3 or one must be a factor of 9. Try a=3 or -3 as the two most likely solutions and a quick check shows that a=3 works. Factoring the (a-3) term give an easily solvable quadratic.
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.
We can also use the sum of roots sum of product of roots and products of roots formulae
i always appreciate how presh calls it brahmagupta's formula [:
Exactly how I was taught to solve these equations in 11th grade. I'm sure in other countries they learn this technique as early as 6th grade, or maybe even earlier!
No way this is a Math Olympiad question, though.
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