Many people in the comments (correctly) pointing out this person used the most convoluted method to reach an answer. Meanwhile I'm going crazy about how this guy writes the letter B and it drives me insane.
They just say that so you'll feel special if you're able to solve it. Same concept behind those ads that show an average difficulty puzzle being solved and say "oNLy pEoPLe WiTh 2o0 iQ cAn SoLvE tHiS". Pure brain rot trash that doesn't deserve our attention.
Before watching: Alright, we can rewrite the initial equation as 3b = b^3. Then factoring out b on both sides gives us 3(b)=(b^2)(b). Before we divide both sides by b, we have to account for the possibility of b=0. If b=0, we have 3(0)=(0^2)(0) -> 0=0. Thus, b=0 is one solution. Having accounted for that, we look for the non zero solutions. We divide both sides by b, and get 3=b^2 From there, we simply square root both sides, and obtain b=sqrt 3 and b= -sqrt 3 as our non-zero solutions.
I taught math for 10 years at a community college. 92% accords with my impression and it is really not so concerning. I am 65 years old and the number of times that I have needed to know how to solve a cubic equation of this sort while performing the activities of everyday life is zero. As a lover of mathematics I would certainly like it if everyone shared my feelings but I need to be realistic about what other people actually want.
I aced my college algebra final. My teacher called and asked if I wanted to pursue a career in math. I said no as I already worked on the railroad. Haven’t used math since basically.
@@rickeyjones729 Your comment is somewhat incoherent to me. Perhaps you could edit it for better understanding. Did common sense get the two roots besides the root of zero?
i thought the same, BUT then, after examining his way, I came to the realization that in our case, we have to actively use the zero in our equation when we are at 3b=b^3, the same time his way didnt have to test numbers, but he came to all solution at once, without testing number i repeat... he found the roots as guided by algebra with the equation of b(b^2-3)=0.... all the solution are in this equation, even zero, without guessing, trying numbers , which is not wrong obviusly, but his way seems more formal to my eyes...
I did the exact same thing: 3b/b = b^3/b -> b^2 = 3. I didn't find out the first solution of 0 to be honest. Which I could have done just brute forcing and approaching the first numbers of both sides of the integer sets. But I was too lazy to write anything down and feared ending up with quadratic equations, which identities' I don't really know that well anymore and I confuse them with each other all the time. And I hate discriminants. Too many symbolic manipulations and calculations for me NOT to fuck up.
Better to impress your math teacher I suppose. Being clever enough to logic out some of the intended complexity is not what they were planning to mark. @g.g.7196
@@g.g.7196 it's not guessing... We see b^3 = 3b and decide to divide by b. Division by 0 isn't allowed, thus our solutions found using this method are only valid for all b unequal to 0. Therefore, we need to check if 0 was a solution, too. There is no guessing involved here.
OK: 3b=b³ => b³-3b = 0 => b*(b²-3) =0 First solution: b=0. Second and third solution: b²=3 => b2/3 = +/- Sqrt(3) Since it is a cubic equation, there are no more solutions.
My first instinct was to divide both sides by b without taking into consideration that b = 0 is a solution. I bet that's where a lot of students trip up: recognizing sqrt3 and maybe -sqrt3 as solutions, but not 0.
@@Contrib92 I didn't divide both sides by 0, I divided both sides by b, giving 3=b^2; read my answer carefully. (b=0 is only one answer to the problem).
It’s easy to get wrong if you miss one or more solutions. If you’re too quickly dismissive, you may miss one or more solutions. I wasn’t paying close attention and solved it quickly in my head before pressing play. Totally missed b=0 as a solution. I’m sure most of the people who missed the question did their math correctly but forgot to recall that cubic equations have 3 roots.
@@DaveJ6515 Doesn’t matter. In this case the solution has 3 roots. I found two and didn’t think to look for a third before clicking the video. Similar issue for anyone who only gets one answer. It isn’t really solved until it’s determined that the other two roots are complex.
Not everyone is math oriented - doesn't make them stupid. Everything is easy when u know it. It is up 2 those of us who R good in math 2 help them learn.
So, I started off by condensing and bringing the b's to one side, giving me b^3-3b=0. Then I factored out b to give me b(b^2-3)=0. This gave me the first solution of b=0. Then divide b from both sides to get b^2-3=0. I then performed the quadratic formula with a=1, b=0, and c=-3. This gave me (-0+/-sqrt(0^2-4(1)(-3)))/2(1) After simplifying, I end up with +/-sqrt(12)/2 sqrt(12) can be split into sqrt(4)*sqrt(3), with sqrt(4)=2. This gave me +/-(2sqrt(3))/2 2's cancel out leaving me with my final two solutions of b=sqrt(3) and b=-sqrt(3) I then plugged them both into a calculator with the original problem and they were equal.
I was crunching it by hand, same logic as you were using, and stopped at (+/-sqrt(12))/2 because I had already 'cheated' once to confirm the layout of the Quatratic Formula (I don't use it often for work, so my memory got fuzzy). So I was just one Commumicative property away from finding the other two real solutions.
why complicate it so much? after your first step of combining like terms, just divide both sides by b. you’ll be left with 3 = b^2. take the sqrt of both sides and you’ll get your +-sqrt3 in one step. the b = 0 is an obvious case that doesn’t require any solving.
It's so easy, divide both sides by b, 3=b², so b=±√3. b=±√3,0. Always think about the equation but substituting values for the variables, if b=1 Then 3=1 is the equation, which is wrong, but also, keep in mind that LHS>RHS If b=2 Then 6=8, again wrong, but this time, LHS
Why did bro do all that? Just do 3b = b^2 × b 3b/b = b^2 3 = b^2 b=+/- √3 So glad i didnt waste 8 mins of my life and just went to see the answer straight away.
If I was a teacher I would have given you a 33 ⅓%. Reason: This is a cubic equation and therefore has 3 solutions. What are the other 2 solutions? In the math world, it is not required to explicitly state to give all 3 solutions. You should know this, if you were taught correctly.
I learned two things. 1) I've forgotten everything from school concerning math. 2) people who didn't are really arrogant about it. Specifically the comments section.
3b = b^3 consider b =\= 0 then we would have 3 = b^2 => b = +- sqrt(3) Now we check if b = 0 is a solution 3(0) = 0^3 => 0 = 0 true Therefore our solutions the roots are +-sqrt(3) and 0
Derp, I solved it as if b was a 3-vector. Same results for each component, but it includes every permutation, giving 27 different real results instead of 3.
Why wouldn't you just divide both sides by b to end up with 3=b squared, which can be reduced to b= +/- square root of 3? Wtf is up with all these extra steps?
Because if you immediately change the equation to 3 = b^2, then you miss the b=0 solution. The way he did it is the most thorough way to prove all solutions were found.
b=b³-2b b=b(b²-2) 1=b²-2 3=b² b=±√3 (0 also works but that's too obvious) ✅ Got sniped by the thumbnail and had to figure it out before clicking, tricksy
Why make it so complicated... 3b=b^3 simplifies to 3=b^2, which is +/- sqrt 3. And the obvious answer stated at the start that can be found from observation and no algebra, zero.
So the conclusion for the college entrance exam question was all three answers were needed to get it right. I can see if students stopped after putting down only one answer, that 100% would have been right but 92% would have been counted as wrong.
Edit: did it that way b4 watching the video, but thats how the guy did it, so yeah b^3 = 3b b^3 - 3b = 0 (b)(b^2 - 3) = 0 Either b = 0 (first solution) or b^2 - 3 = 0 b^2 = 3 b = +-sqrt(3) (second and third solutions)
@@DeliberateContrarian the equation is 3b = b^3 then you divide by b and get: 3 = b^2 then you do a square root and get this: b = √ 3 then it must be that √ 3+√ 3+√ 3 = √ 3*√ 3*√ 3 left side is 3*√ 3 because there are three √ 3 and right side is 3*√ 3 because √ x*√ x = x and then there is aditional √ x
I divided the left side by b, got: (b+b+b)/b = b^2 then I broke up that fraction into: b/b + b/b + b/b = b^2 that's equal to: 1 + 1 + 1 = b^2 or b^2 = 3 and then solved for b. I got lucky lol.
That ignores the 0 solution. The method shown will always find all the solutions, this is more important when the other solution of 0 is not so obvious.
@@andrewness Yes, but if you're doing it for an exam, or to impress an interviewer, you should be able to demonstrate you can do it for non-trivial cases. You certainly need to point out that by dividing both sides by b, you are losing a solution by converting a cubic to a quadratic, otherwise you would not be showing proper understanding of the problem.
A cubicle equation can be written as b^3 - 3 b = 0
b (b^2 - 3 ) = 0
b = 0 & b = + - √3
Yup, my solution.
Wow found the moron
Did it in my head while looking at the thumbnail.
I was already wondering why Substrat 3b when devidint by b puts the answer in plain sight
@@DarkSider667 basically, one shouldn't go to divide with 0. That's a mathematical fallacy.
Many people in the comments (correctly) pointing out this person used the most convoluted method to reach an answer. Meanwhile I'm going crazy about how this guy writes the letter B and it drives me insane.
☘️
Kinda beautiful tbh looks kinda slavonic
That's why he got it wrong at first. He turned the b into another letter and made it impossible to solve for b.
I can't concentrate enough on what he's trying to explain because my brain keeps screaming, "Upside-down letter g".
3b = b * b * b
3 = (b * b * b) / b
3 = b * b
b = sqrt( 3 )
A much easier way to solve is:
b + b + b = b * b * b
3b = b^3
3 = (b^3)/b
3 = b^2
+/-sqrt(3) = b
(The third answer is 0)
Quicker and simpler way, we must remember that b = 0 too :)
Spot on - exactly how I did it
@@jocelynayub5027 Me too.
This also was the way I solved this (75 year old non-college), a lot of unnecessary extra work when setting to zero
Well, you have to check b=0 first, which is trivial, before you can divide both sides by b... Since that implies b doesn't equal 0. But yeah ;)
the fact that 92% got it wrong is concerning
Maybe because they got one answer instead of 3
@@warmth9140yep, that’s why I would have gotten it wrong. Literally only simplified the equation to get my plus/minus root 3 and forgot 0 🤦♂️
Yes. Very concerning
glancing at the thumbnail I only got 2 roots, +(3)^.5 and -(3)^.5.. i forgot 0 in my haste. very concerning indeed.
They just say that so you'll feel special if you're able to solve it. Same concept behind those ads that show an average difficulty puzzle being solved and say "oNLy pEoPLe WiTh 2o0 iQ cAn SoLvE tHiS".
Pure brain rot trash that doesn't deserve our attention.
Before watching:
Alright, we can rewrite the initial equation as 3b = b^3. Then factoring out b on both sides gives us
3(b)=(b^2)(b).
Before we divide both sides by b, we have to account for the possibility of b=0. If b=0, we have 3(0)=(0^2)(0) -> 0=0.
Thus, b=0 is one solution.
Having accounted for that, we look for the non zero solutions. We divide both sides by b, and get 3=b^2
From there, we simply square root both sides, and obtain b=sqrt 3 and b= -sqrt 3 as our non-zero solutions.
I hope that 92% metric is pure BS, because college students not being able to solve this simple problem is highly concerning.
100% of Art majors, russian literature, English, poly sci majors
I taught math for 10 years at a community college. 92% accords with my impression and it is really not so concerning. I am 65 years old and the number of times that I have needed to know how to solve a cubic equation of this sort while performing the activities of everyday life is zero. As a lover of mathematics I would certainly like it if everyone shared my feelings but I need to be realistic about what other people actually want.
I aced my college algebra final. My teacher called and asked if I wanted to pursue a career in math. I said no as I already worked on the railroad. Haven’t used math since basically.
This problem u didn't need an equation though common sense try 1 and then u automatically knew it was 0 since b is the same number just looking at it.
@@rickeyjones729 Your comment is somewhat incoherent to me. Perhaps you could edit it for better understanding. Did common sense get the two roots besides the root of zero?
Writing 3b=b^3 give first root b=0. If b#0, let's divide: 3=b^2; no more than 2 solutions, +-sqrt(3). One line
i thought the same, BUT then, after examining his way, I came to the realization that in our case, we have to actively use the zero in our equation when we are at 3b=b^3, the same time his way didnt have to test numbers, but he came to all solution at once, without testing number i repeat... he found the roots as guided by algebra with the equation of b(b^2-3)=0.... all the solution are in this equation, even zero, without guessing, trying numbers , which is not wrong obviusly, but his way seems more formal to my eyes...
I did the exact same thing: 3b/b = b^3/b -> b^2 = 3. I didn't find out the first solution of 0 to be honest. Which I could have done just brute forcing and approaching the first numbers of both sides of the integer sets. But I was too lazy to write anything down and feared ending up with quadratic equations, which identities' I don't really know that well anymore and I confuse them with each other all the time. And I hate discriminants. Too many symbolic manipulations and calculations for me NOT to fuck up.
Better to impress your math teacher I suppose.
Being clever enough to logic out some of the intended complexity is not what they were planning to mark.
@g.g.7196
@@g.g.7196 it's not guessing... We see b^3 = 3b and decide to divide by b. Division by 0 isn't allowed, thus our solutions found using this method are only valid for all b unequal to 0. Therefore, we need to check if 0 was a solution, too. There is no guessing involved here.
OK: 3b=b³ => b³-3b = 0 => b*(b²-3) =0 First solution: b=0.
Second and third solution: b²=3 => b2/3 = +/- Sqrt(3)
Since it is a cubic equation, there are no more solutions.
Clearly, the first root is b=0; but, given that 3b=b³, then dividing both sides by b gives --> 3=b², hence b=+-Sqrt(3). It's not rocket science!!
My first instinct was to divide both sides by b without taking into consideration that b = 0 is a solution. I bet that's where a lot of students trip up: recognizing sqrt3 and maybe -sqrt3 as solutions, but not 0.
@@hls6925you can't solve this way because you can't divide by 0
@@Contrib92 I didn't divide both sides by 0, I divided both sides by b, giving 3=b^2; read my answer carefully. (b=0 is only one answer to the problem).
This is straight forward, nothing complicated. 2 steps and it's done.
This is NOT a tricky question. It’s a ridiculously easy question.
Once I wrote it down as bxbxb = 3xb it was easy to see. But the original presentation led me in the wrong direction at first.
Just eliminate one b by the division, and you're left with 3 = b², and the rest is obvious enough.
Skipped over 0 being another solution
@@TheRealSmilerVision That was just one alternative answer, the other ones were indeed ±√3
@alfianfahmi5430 then we agree
Yes I’m confused by the length. I did it in my head in two lines?
@@frankslade33yes!!!!!!
3*b = b^3
You can divide both sides by b, therefore (b = 0) is a solution
3 = b^2
b = [+/-]sqrt(b)
Solutions...
b= 0 ; sqrt(b) ; -sqrt(b)
Easy it can be solved by 9th grade
It’s easy to get wrong if you miss one or more solutions. If you’re too quickly dismissive, you may miss one or more solutions. I wasn’t paying close attention and solved it quickly in my head before pressing play. Totally missed b=0 as a solution. I’m sure most of the people who missed the question did their math correctly but forgot to recall that cubic equations have 3 roots.
@@zemoxianin R either 1 or 3.
√3
@@DaveJ6515
Doesn’t matter. In this case the solution has 3 roots. I found two and didn’t think to look for a third before clicking the video. Similar issue for anyone who only gets one answer. It isn’t really solved until it’s determined that the other two roots are complex.
Not everyone is math oriented - doesn't make them stupid. Everything is easy when u know it. It is up 2 those of us who R good in math 2 help them learn.
3b=b^3
3b-b^3 = 0
b(3-b^2)=0
b(sqrt3 - b)(sqrt3 + b) = 0
b = 0 or b = -(sqrt3) or b = sqrt 3
Before watching the video:
I’m locking in my answer as 0
Damn so yeah we have 3 roots: -r, 0 and r for a specific value r.
So many replies suggesting dividing both side by "b" is scary to me due to the risk of "losing" the 3rd root to this cubic.
So, I started off by condensing and bringing the b's to one side, giving me b^3-3b=0. Then I factored out b to give me b(b^2-3)=0. This gave me the first solution of b=0. Then divide b from both sides to get b^2-3=0. I then performed the quadratic formula with a=1, b=0, and c=-3. This gave me (-0+/-sqrt(0^2-4(1)(-3)))/2(1) After simplifying, I end up with +/-sqrt(12)/2 sqrt(12) can be split into sqrt(4)*sqrt(3), with sqrt(4)=2. This gave me +/-(2sqrt(3))/2 2's cancel out leaving me with my final two solutions of b=sqrt(3) and b=-sqrt(3) I then plugged them both into a calculator with the original problem and they were equal.
I was crunching it by hand, same logic as you were using, and stopped at (+/-sqrt(12))/2 because I had already 'cheated' once to confirm the layout of the Quatratic Formula (I don't use it often for work, so my memory got fuzzy). So I was just one Commumicative property away from finding the other two real solutions.
This is how we solve the problem in an exam to get good marks
Why would you choose the letter ‘b’ in the formula if your hand written letter B is that horrendous???
Surely you mean that YOUR way to hand-write a "b" is horrendous?
@@Scott-i9v2s Perhaps, and don't call me Shirely
@@lowrez-v3d 🙂 (caught me, fair-&-square!)
No need for "difference of squares" formula.
b^2 = 3 gives immediately
b = ±√3
I only got 2 roots cuz i basically forgot about finding 3 roots for 3 degree equation 😅
Same. I solved for the roots in three steps, and never noticed the painfully obvious "0" solution that my method missed. 😅
why complicate it so much? after your first step of combining like terms, just divide both sides by b. you’ll be left with 3 = b^2. take the sqrt of both sides and you’ll get your +-sqrt3 in one step. the b = 0 is an obvious case that doesn’t require any solving.
Oh my! I like the way he took the longest route. Just a you tube instinct.
I would have flunked out if I made a "b" like he does.
Ya , his “ b “ looked like a curly 6
And sometimes an 8.
It's so easy, divide both sides by b, 3=b², so b=±√3.
b=±√3,0.
Always think about the equation but substituting values for the variables, if b=1
Then 3=1 is the equation, which is wrong, but also, keep in mind that LHS>RHS
If b=2
Then 6=8, again wrong, but this time, LHS
My first intuition was to just divide each side by b.
3 = b^2 => b = √3
3b=b^3
3b/b=b^2
b^2=3
b=+/-sqrt(3),0
Simple. Bro is onto cooking nothing.
Why did bro do all that?
Just do 3b = b^2 × b
3b/b = b^2
3 = b^2
b=+/- √3
So glad i didnt waste 8 mins of my life and just went to see the answer straight away.
And thus you overlooked b=0 as valid root...
The answer is 0. It works in the beginning.😂😂😂
If I was a teacher I would have given you a 33 ⅓%.
Reason:
This is a cubic equation and therefore has 3 solutions.
What are the other 2 solutions? In the math world, it is not required to explicitly state to give all 3 solutions. You should know this, if you were taught correctly.
@@topkatz58 There are answers and then complete answers. But if the question just wants ONE answer, then 0 works.
@@topkatz58 Solution 2 - answer is nil. Solution 3 - answer is nought.
@@topkatz58 thank God u aren't a teacher
Ok, but remember if you had something in the real world you wasted to solve for, the positive srqt of 3 makes more sense.
All those bees and no pollination, shame...
It concerns me the amount of people who believe this can be solved dividing by 0
Isn’t one side of the equation addition and the other side multiplication?
ax=x^b
Very long winded method.
This isn't hard, you just did the most convoluted method of solving to make it appear complicated.
A simpler way is just to divide both sides by b. 3b=b^3. (3b)/b = (b^3)/b. 3 = b^2.
I learned two things. 1) I've forgotten everything from school concerning math. 2) people who didn't are really arrogant about it. Specifically the comments section.
Right on.
3b = b^3 consider b =\= 0 then we would have
3 = b^2 => b = +- sqrt(3)
Now we check if b = 0 is a solution
3(0) = 0^3 => 0 = 0 true
Therefore our solutions the roots are +-sqrt(3) and 0
Not sure why this was even a challenge for any proper math student past 11th grade
Good video. Question: Did you originally write with your left hand and then transition to the right hand?
8 minutes for a 30 seconds problem.
Obvious answer b=0
Then looking for solutions =/= 0 we can divide by b
b²=3 b is +/- sqrt(3).
I hate it there was no youtube when i was doingg this at school
0, +sqrt(3), -sqrt(3)
Zero!
Yup darn it I forgot to consider all the fancy polynomial things I rarely use.
Finally, one of these that only took me 15 seconds for a change. b(b+sqrt(3))(b-sqrt(3))=0
Derp, I solved it as if b was a 3-vector. Same results for each component, but it includes every permutation, giving 27 different real results instead of 3.
b+b+b=b.b.b => 3b=b^3 => (3b)/b=(b^3)/b => 3=b^2
Why wouldn't you just divide both sides by b to end up with 3=b squared, which can be reduced to b= +/- square root of 3? Wtf is up with all these extra steps?
Because if you immediately change the equation to 3 = b^2, then you miss the b=0 solution. The way he did it is the most thorough way to prove all solutions were found.
Because b could be zero, and division by zero is a no-no.
One can also divide both sides by b to get 3=b², after which we get b=±Sqr(3).
b = 0, \quad b = \sqrt{3}, \quad b = -\sqrt{3}
Divide both sides by 3.
3 = b×b.
1) b^3 = 3b
b^3/b =3
b^2 =3
b = rt(3) , b= 3i , b=0
b=b³-2b
b=b(b²-2)
1=b²-2
3=b²
b=±√3 (0 also works but that's too obvious)
✅
Got sniped by the thumbnail and had to figure it out before clicking, tricksy
I just divided both sides by b and got 3=b^2 took square root both sides and got b=+-sqrt(3) and then there’s 0 obviously
3b=b³
b³-3b=0
b(b²-3)=0
b=0 or b=±sqrt(3)
this is like a 10second problem to solve
b=0, b=sqrt(3), b=-sqrt(3)
Got it wrong but I did understand how it easy it really was but my brain wasn't functioning 😭
b+b+b = b.b.b
3b = b^3
b(b^2-3) = 0
=> [b = 0
[b^2 - 3 = 0 => [b = + sqr(3)
[b = - sqr(3)
Why didn't you just divide the original equation by b to get 3=b^2 ?
Yes but b =0 can be a thing
And u can't divide by 0
Why make it so complicated... 3b=b^3 simplifies to 3=b^2, which is +/- sqrt 3. And the obvious answer stated at the start that can be found from observation and no algebra, zero.
You can solve it with guess-and-check in like 10 seconds. What do you even need algebra for?
Either 0, 0 and 0, or 1, 2 and 3.
You coukd just do 3b/b = b³/b to get 3 = b² and then b = sqrt(3)
Love the way you write.
Excellent!
So the conclusion for the college entrance exam question was all three answers were needed to get it right. I can see if students stopped after putting down only one answer, that 100% would have been right but 92% would have been counted
as wrong.
b + b + b = b • b • b
3b = b³
b³ - 3b = 0
(b²-3)b = 0
b² - 3 = 0 | b = 0 ✓
b² - (√3)² = 0
(b+√3)(b-√3) = 0
b + √3 = 0 | b - √3 = 0
b = -√3 ✓ | b = √3 ✓
[ b = -√3, 0, √3 ]
Edit: did it that way b4 watching the video, but thats how the guy did it, so yeah
b^3 = 3b
b^3 - 3b = 0
(b)(b^2 - 3) = 0
Either b = 0 (first solution) or b^2 - 3 = 0
b^2 = 3
b = +-sqrt(3) (second and third solutions)
3b = b^3
0 = b^3 - 3b
0 = b(b^2 - 3)
b = 0 or 0 = b^2 - 3
3 = b^2
b = root3 or -root3
0, and +root3 and -root3
b=0
0+0+0=0
0*0*0=0
I may be dumb because I only saw the sqrt3 solution but not the obvious 0 solution
You can easily get the √3 and -√3 by simply solving an equation and then there is 0 because anything * 0 = 0
I don't understand how the square root of 3 plus the square root of 3 is 3.
@@DeliberateContrarian the equation is 3b = b^3
then you divide by b and get: 3 = b^2
then you do a square root and get this: b = √ 3
then it must be that √ 3+√ 3+√ 3 = √ 3*√ 3*√ 3
left side is 3*√ 3 because there are three √ 3 and right side is 3*√ 3 because √ x*√ x = x and then there is aditional √ x
interesting! also, what's the pen you're using?
Cubic function. 3 roots by FTA.
Failing to recognize the easiest solution is the wrong answer in my opinion. b^3/b => b^2
The square root of 3 was my immediate answer
(1)+(1)+(1)=3 (3)*(3)*(3)=27 3^3(b ➖ 3ib+3i).
b+b+b =b×b×b
3b =b³
b³ -3b =0
b(b² -3) =0
☆b =0
b² -3 =0
(b +3⁰•⁵)(b -3⁰•⁵) =0
☆b =±3⁰•⁵
Are you kidding me?
3b=b^3; b^2=3; b=3^0.5. Then fiddle for any other solutions.
I'm sure you'll understand if I block this drivel.
I forgot about 0, but I also hate square roots
3b - b³ = 0
Solved this in my head before watching 👍
Same with no formula just knew 0 times itself and added together was 0. Lol
Can you demonstrate Σb = Πb and unpack the deeper insight?
I divided the left side by b, got:
(b+b+b)/b = b^2
then I broke up that fraction into:
b/b + b/b + b/b = b^2
that's equal to:
1 + 1 + 1 = b^2 or b^2 = 3
and then solved for b. I got lucky lol.
0, √3 and -√3
Is it weird that my first thought was to just simplify it and divide both sides by b? 3 = b^2 .... So b=+-√3
Ok just by looking at this problem I see it as b=0.
It is not a "Math Problem" as much as a "Logic Exercise"
And thus you overlooked the other 2 roots...
3b=b^3
so b^3-3b=0
so b=0 or b^2=3
Didn't do it that way. I just divided both sides (3b =b cubed) by b to get 3 = b squared. Seemed partly correct to me.
b x (b^2-3)=0
b = (b+b+b) / b^2
0 right!??!?!?!
Excellent
i solved this when i was sitting in my bathroom without calculator and paper 😭🙏
0 or ± sqrt 3
Seems very complicated method when there's a simpler way.
b+b+b=3b
So 3b=b x b x b
Divide both sides by b
3=b x b
b is root 3.
That ignores the 0 solution. The method shown will always find all the solutions, this is more important when the other solution of 0 is not so obvious.
@MentalLentil-ev9jr solving for 0 is trivially easy in this case though
@@andrewness Yes, but if you're doing it for an exam, or to impress an interviewer, you should be able to demonstrate you can do it for non-trivial cases. You certainly need to point out that by dividing both sides by b, you are losing a solution by converting a cubic to a quadratic, otherwise you would not be showing proper understanding of the problem.
sqrt of 3 solves it. Don't tried further to see if there are more solutions.
Now I will watch the videos
Dude
It’s zero
Obviously
That's 3b = b³ => 3b - b³ = 0 => b (3 - b²) = 0 so either b=0 or 3-b²=0 => b²=3 => b=√3
Solved ez
3b = b³
3 = b²
b = ±√3
3b = b³
3b - b³ = 0
b(3 - b²) = 0
b = 0
So if you had 4 bs on both side would it be sqrt 4?