I just used x=u³ in the expression and got: 4root(u³ + 15) = u + 1 Then elevated both sides to the 4th power: u³ + 15 = u^4 + 4u³ + 6u² + 4u + 1 => u^4 + 3u³ + 6u² + 4u - 14 = 0 => (u - 1)(u³ + 4u² + 10u + 14) = 0 And it was at that point I thought I was doing everything the hard way, because that 3rd degree polynomial didn't have "nice" roots, so I gave up and started watching the video - that's when I discovered this problem isn't "nice" at all. ¯\_(ツ)_/¯
Wow, nice similar. I had quite similar, but I didn't know how to find a solution of the cubic equation, so I proved there is only one solution using a derivative, then I estimated it to be between -3 and -2 and then I used Newton's method. I'm not accustomed to engaging a formula for the root of a cubic equation.
The second value of b leads to an extraneous solution . The value of x that it yield leads to a positive value on the left hand side of the original equation and a negative value on the right hand side of the original equation . This extraneous solution happens because in part of your solution you raise (b+1) to the fourth power .
6:55 Here's an easy way to handle the cubic term , just substitute b = a - 1 in the cubic polynomial , you will get ( a^3 + a^2 + 5a + 4 ) . Now , since *a* is a 4th root , hence a >= 0 which means a^3 + a^2 + 5a + 4 >= 0 + 0 + 0 + 4 = 4 > 0 , therefore the cubic term never equal to 0 hence the only solution is x = 1 :)
Note :--- I think your second solution in the video is incorrect because to satisfy the domain in sqrt(sqrt(x+15)) , x >= - 15 . But your second solution is less than - 15 .
⁴√(x + 15) = ³√x + 1 Substitute x = u³: ⁴√(u³ + 15) = u + 1 Raise to 4th power: u³ + 15 = u⁴ + 4u³ + 6u² + 4u + 1 Move all terms to LHS and negate: u⁴ + 3u³ + 6u² + 4u - 14 = 0. As the sum of the coefficients is zero, u - 1 is a factor. Dividing through, (u - 1)|(u⁴ + 3u³ + 6u² + 4u - 14) = (u³ - 4u² + 10u - 14) (u - 1)(u³ - 4u² + 10u - 14) = 0 u = 1 or u³ - 4u² + 10u - 14 = 0. However, this cubic has no nice factorisation, so we'll have to use the cubic formula or substitution to solve it. Wolfram Alpha suggests substituting 'u = v + 4/3'. I'm not going to solve it through here, but there's only one real solution to the cubic with a really messy exact form. So lets take u = 1. Now, x = u³ so: x = 1.
X is also equal to approx -12.167 ... the other value (-12.167)^1/3 = -2.3 + 1= -1.3 = absolute value = 1.3, calculation was for the right side proof. now let us proof check the left side: (-12.167 + 15)^1/4 = (2.833)^1/4 = approx. 1.297... absolute(1.3) approx. 1.297 -2.3 plugs into the cubic equation converges close to zero... u^3+4u^2+10u +14=0.... (-2.3)^3 = -12.167
@@ivandebiasi6657 That's impossible. We can easily prove, that there are no other solutions. Only solution is x = 1. We can take a limit (x approaching infinity) [(x+15)^1/4]/[(x)^1/3 + 1]. If this limit will be greater than one, it will mean, that the top function is eventualy bigger (in infinity). And this limit is equal to infinity, which is obviously bigger than one, so that means, there is no other intersection point.
your 2nd solution *x=-12.1521* (approx. value) results in *1.29906* for the left-hand side of your problem statement (positive!) and *-1.29906* for the right-hand side (negative!). in addition, wolframalpha shows 1 solution only, enter on the website. I am confident that there is only 1 solution, namely x=1: *𝚁𝚎𝚍𝚞𝚌𝚎[𝚂𝚞𝚛𝚍[𝟷𝟻 + 𝚡, 𝟺] == 𝚂𝚞𝚛𝚍[𝚡, 𝟹] + 𝟷, 𝚡]*
I think may be the second solution does not actuallly satify the equation.It satifies the negative fourth root rather than the positive. But I could be wrong.
@@SyberMath I don't see this as a mistake but more a matter of convention, which to be honest I am not all that sure about. I think the idea is that these root symbols are intended to be the principle branch but I am not sure about this. There are situations when you most definitely don't want to do this. I learn a lot from your videos and find they have improved me mathematically.
The only real solution of b^3+4b^2+10b+14=0 is b=-2.29906. Thus b+1 is negative whilst (15+b^3)^1/4 is positive. Therefore this value of b does not satisfy the original equation. The only solution (real) is b=1 > x=1.
I only found x=1 as a solution and I think I did it right(its good not to find the other solution since its ugly and perhaps don't satisfy the equation)
I just used x=u³ in the expression and got:
4root(u³ + 15) = u + 1
Then elevated both sides to the 4th power:
u³ + 15 = u^4 + 4u³ + 6u² + 4u + 1
=> u^4 + 3u³ + 6u² + 4u - 14 = 0
=> (u - 1)(u³ + 4u² + 10u + 14) = 0
And it was at that point I thought I was doing everything the hard way, because that 3rd degree polynomial didn't have "nice" roots, so I gave up and started watching the video - that's when I discovered this problem isn't "nice" at all. ¯\_(ツ)_/¯
Nice!
pretty good try lol
Wow, nice similar. I had quite similar, but I didn't know how to find a solution of the cubic equation, so I proved there is only one solution using a derivative, then I estimated it to be between -3 and -2 and then I used Newton's method. I'm not accustomed to engaging a formula for the root of a cubic equation.
Don't worry man, no one is used to that cubic formula, it's really horrible.
The second value of b leads to an extraneous solution . The value of x that it yield leads to a positive value on the left hand side of the original equation and a negative value on the right hand side of the original equation . This extraneous solution happens because in part of your solution you raise (b+1) to the fourth power .
What @pk2712 is saying is that this video is completely wrong.
6:55 Here's an easy way to handle the cubic term , just substitute b = a - 1 in the cubic polynomial , you will get ( a^3 + a^2 + 5a + 4 ) . Now , since *a* is a 4th root , hence a >= 0 which means a^3 + a^2 + 5a + 4 >= 0 + 0 + 0 + 4 = 4 > 0 , therefore the cubic term never equal to 0 hence the only solution is x = 1 :)
Note :--- I think your second solution in the video is incorrect because to satisfy the domain in sqrt(sqrt(x+15)) , x >= - 15 . But your second solution is less than - 15 .
Uh oh! 😂
⁴√(x + 15) = ³√x + 1
Substitute x = u³:
⁴√(u³ + 15) = u + 1
Raise to 4th power:
u³ + 15 = u⁴ + 4u³ + 6u² + 4u + 1
Move all terms to LHS and negate:
u⁴ + 3u³ + 6u² + 4u - 14 = 0.
As the sum of the coefficients is zero, u - 1 is a factor.
Dividing through,
(u - 1)|(u⁴ + 3u³ + 6u² + 4u - 14) = (u³ - 4u² + 10u - 14)
(u - 1)(u³ - 4u² + 10u - 14) = 0
u = 1 or u³ - 4u² + 10u - 14 = 0.
However, this cubic has no nice factorisation, so we'll have to use the cubic formula or substitution to solve it. Wolfram Alpha suggests substituting 'u = v + 4/3'. I'm not going to solve it through here, but there's only one real solution to the cubic with a really messy exact form. So lets take u = 1. Now, x = u³ so:
x = 1.
this one was complex, great job bro, watching a master
Thank you!
X is also equal to approx -12.167 ... the other value (-12.167)^1/3 = -2.3 + 1= -1.3 = absolute value = 1.3, calculation was for the right side proof.
now let us proof check the left side: (-12.167 + 15)^1/4 = (2.833)^1/4 = approx. 1.297... absolute(1.3) approx. 1.297
-2.3 plugs into the cubic equation converges close to zero... u^3+4u^2+10u +14=0.... (-2.3)^3 = -12.167
If you graph both functions, they intersect only once. Therefore I think, there is only one real solution.
agree the other solution does not work
That's right!
There is another solunzion that is near 8.2749 according to Geogebra... but it isn't the one you found
@@ivandebiasi6657 That's impossible. We can easily prove, that there are no other solutions. Only solution is x = 1. We can take a limit (x approaching infinity) [(x+15)^1/4]/[(x)^1/3 + 1]. If this limit will be greater than one, it will mean, that the top function is eventualy bigger (in infinity). And this limit is equal to infinity, which is obviously bigger than one, so that means, there is no other intersection point.
The other solution was extraneous and was brought by the substitution and raising it to the fourth power. Hence x=1 is the only real solution.
How can I learn about the second solution? I was completely lost on how that was created. Anyone have any links?
Second solution does not satisfy the original equation
b^3+4b^2+10b+14=(2b+5/2)^2+31/4+b^3 > 0+31/4-1=27/4>0, therefore only x = 1
Левая часть неотрицательна.
Значит и правая часть тоже.
squb x+1>=0. x>=-1.
Только один корень x=1
That's right!
Fourth 😁
Nice one....wait you uploaded the video so you were first the entire time ?!!
Thanks. I did upload it but was the fourth person to come to the premiere I guess
The complicated solution must be a false root. a must be non-negative but b+1 would be negative.
That's right!
your 2nd solution *x=-12.1521* (approx. value) results in *1.29906* for the left-hand side of your problem statement (positive!) and *-1.29906* for the right-hand side (negative!).
in addition, wolframalpha shows 1 solution only, enter
on the website.
I am confident that there is only 1 solution, namely x=1:
*𝚁𝚎𝚍𝚞𝚌𝚎[𝚂𝚞𝚛𝚍[𝟷𝟻 + 𝚡, 𝟺] == 𝚂𝚞𝚛𝚍[𝚡, 𝟹] + 𝟷, 𝚡]*
Cardano´s solution formula is so complicated, so very few students challenging for entrance exams can
figured out non-integer solutions.
Right! It's not too bad, though
I think may be the second solution does not actuallly satify the equation.It satifies the negative fourth root rather than the positive. But I could be wrong.
You're right!
@@SyberMath I don't see this as a mistake but more a matter of convention, which to be honest I am not all that sure about. I think the idea is that these root symbols are intended to be the principle branch but I am not sure about this. There are situations when you most definitely don't want to do this. I learn a lot from your videos and find they have improved me mathematically.
Glad to hear!
The only real solution of b^3+4b^2+10b+14=0 is b=-2.29906. Thus b+1 is negative whilst (15+b^3)^1/4 is positive. Therefore this value of b does not satisfy the original equation. The only solution (real) is b=1 > x=1.
I really love your videos man :)
Glad you like them!
I got one of the solutions. Another great explanation, SyberMath!
Yeah, me too. I couldn't for the life of me find the x = 1 solution.
😁
Me: looks to complicated, so I just plug in 1 and look if it's right
Also me: lol
👍😁
I only found x=1 as a solution and I think I did it right(its good not to find the other solution since its ugly and perhaps don't satisfy the equation)
agree
Could you please show us in detail how to get second solution of b?
That second solution makes the baby Jesus cry.
Great video. I actually checked and the second solution is not valid.
As always amazing problem 😍
Thank you so much 😀
Woah!one of the solution is very radical
Radical and complex! 😁
please I want to know the general form of the cubic equation
7:29 When he wrote another b solution, people probably lost at it
Whats the point if "wolfram alpha" can do it all for you?
Wow! One nice solution and one horrifying one.
Actually, the radical solution does not satisfy the equation! 😁
@@SyberMath Actually, I was just noticing that, looking at the graphs!
Hmm there's a very simple way I mean very simple standard way and I am gonna make video on it 😁😁
x= 1 has a trivial solution to this equation .
Second
only x=1 is a solution as the other is negative which is incompatible with the sqrt(x) in the initial equation
could not get this one, very complicated solution
make: [(x)^1/3] +1 = z ----> x=1
The second answer was so beautiful
Damn you're so good.....
that's what she said
Couldn't catch the premiere unfortunately
No worries! 🙂
Great style, always make it sound easy, it suppose to be easy😊 easy wurks...
Thank you! 😊
WTH wolframalpha !!!!!!
Came here to say this. If WolframAlpha was an option the whole time, why not use it on the initial equation? 😂
@@tannerstrickler7352 what can we do otherwise? Cubic formula?
@@Noname-67 The cubic formula would be the only way that I know of to solve it by hand, but it's pretty nasty.
Pero amb Geogebra les dues funcions només tenen un punt de tall.
That's right! I made a mistake
The second time i solve it by the eyes
so good
There is a mistake ...there is one solution which is 1 ...the other one is impossible because x €(-1,+infini (
That's right!
I like kind of this equation
Per el mètode de Regula Falsi la segona solució és -2,299062577
😳...just amazing
Thank you!
only one real solution
That's right!
I like you
Thanks!!!!!?!? So much
i dont understand solve degree 3
Please simplify the other solution sybermath 😂😂😂😂
😁
The other solution was so irritating
You know
Wow that’s interested sir
b=-2,299062577
First
8th😂
379th. : )
😁
Bu soru çok saçma bı soru böyle çözüm olamaz.
Why? 🤔
Fifth
Third