I note that the conjugate of sqrt(5)+2 is 2-sqrt(5), not sqrt(5)-2, because the radical changes when you conjugate. My idea for solving this is to add the conjugates together: (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3). This looks like the solution of a cubic equation; in fact, if you "antisolve" the expression (a+sqrt(b))^(1/3) + (a-sqrt(b))^(1/3), by comparing a and b with -q/2 and q^2/4+p^3/27, you get the equation x^3-3*(a^2-b)^(1/3)* x-2a=0. Substituting 2 and 5 for a and b gives x^3+3x-4=0. Lo and behold, 1 solves this equation. The other two roots are complex and can't be right since everything here is real. Therefore (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3) = 1. Now multiply them. (2+sqrt(5))^(1/3) * (2-sqrt(5))^(1/3) = (4-5)^(1/3) = -1. So (2+sqrt(5))^(1/3) and its conjugate satisfy x^2-x-1 = 0, from which you get phi and -1/phi as roots. So (2+sqrt(5))^(1/3) = phi.
Here is the general method I use when trying to denest a cube root radical with a square root inside. Begin by assuming the radical denests as eq1: (A + sqrt(B))^(1/3) = X + sqrt(Y) which if it does hold, then it can be shown (but I will not do so here) that eq2: (A - sqrt(B))^(1/3) = X - sqrt(Y) Now, define the product of the left hand sides of eq1 and eq2 as R R = (A^2 - B)^(1/3) and note that R must be a rational number for the cube root to denest. That is a necessary but not sufficient condition for denesting to succeed. Next, multiply eq1 and eq2 to get eq3: X^2 - Y = R which will be used later. Now add eq1 and eq2 to get 2*X = (A + sqrt(B))^(1/3) + (A - sqrt(B))^(1/3) and cube that to get 8*X^3 = A + sqrt(B) + A - sqrt(B) + 3*R*((A + sqrt(B))^(1/3) + (A - sqrt(B))^(1/3)) = 2*A + 3*R*(2*X) which can be rearranged to standard form for a depressed cubic equation eq4: 4*X^3 - 3*R*X - A = 0 which, if you try to solve using Cardano's formula you will just get more nested radicals. So use the rational root theorem to find a root X1: (any factor of A) / (± any factor of 4) There are 6 possibilities for the denominator (±1, ±2, ±4) and N number of possibilities for the numerator (do not double for plus/minus for the numerator or you double count) so at most 6*N for the total number of possible rational roots to try in eq4 (usually fewer depending on the actual factors). Of course you can stop once you find one that works. Then you can determine Y from eq3 Y = X^2 - R In this case, A = 2 , B = 5, and R = -1 so the cubic is 4*X^3 + 3*X - 2 = 0 so there are at most 6*2 = 12 possibilities for a rational root as (1,2) / (±1, ±2, ±4) but there are actually only 8 possibilities in this case and I find that X = 1/2 is a root, and then Y = X^2 - R = 1/4 - (-1) = 5/4 so we have X + sqrt(Y) = 1/2 + sqrt(5/4) = (1 + sqrt(5)) / 2
@Afnaan2701 All cubic formulas such as Cardano's formula will just give you a nested radical, which is what we already are struggling with. That is why you need to use the rational root theorem instead of a formula (or the method used to derive the formula, which is of course the same as using the formula)
@Afnaan2701 No, that is incorrect. As I said, there are formulas for the solution to cubic equations, such as Cardano's formula. The issue is that the formula contains nested radicals, so the formulas are generally not helpful in denesting nested radicals.
Or if you know that phi = (root(5)+1)/2, you can note that root(5) + 2 = root(5) + 1 + 1 = 2 phi +1, and use phi^2 = phi +1 to see that phi^3 = phi^2 + phi = 2 phi +1, so cuberoot(2 phi +1) = cuberoot(phi^3) = phi.
Here's a general method, with no "magic" or "deus ex-machina" in it, although the final step does involve an obvious guess. Since radicals and integers are independent bases, we know that ∛(√5+2) will be of the form a + b√5. Cubing (a + b√5), we get a^3 + 3a^2*b√5 + 15a*b^2 + 5√5*b^3. Equating coefficients of the integers of that and (√5+2), we get a^3 + 15a*b^2 = 2. Equating coefficients of the √5 parts gives 3a^2*b + 5b^3 = 1. Now, those look rather difficult to solve for a and b, but you can see a solution more easily by substituting c=2a and d=2b, then multiplying both sides of both equations by 8. That gives 3c^2*d + 5d^3 = 8 and c^3 + 15c*d^2 = 16. Looking at the coefficients indicates c=1, d=1. So a=1/2, b=1/2 and therefore ∛(√5+2) = 1/2 + √5/2.
u + 1/u = √5 provides 2 solutions u1 and u2 since u1 is = 1/u2 and viceversa. Both are positive and one is greater than 1. The other, since it's the reciprocal of tht one, is less than 1. We need the greater one. So, pick the + version
@@leif1075 Thanks for the question! Someone who dealt with radical expressions for while and solved a good variety of problems on radicals should know how to use conjugates in certain situations. It takes practice and willingness to learn these. I know it's not always very straightforward and that's why I presented an alternative solution method.
With a little bit of Pell Equation theory one can just write: √5 + 2 = ⅛⋅(5√5 + 15 + 3√5 + 1) = ⅛⋅(5√5 + 3⋅5 + 3⋅√5 + 1) = ⅛⋅(√5³ + 3⋅√5² + 3⋅√5 + 1) = (½)³⋅(√5 + 1)³. This may look magical, but is really just recognizing the Pell Eqn consequences of the second method; and applying them. One knows to use 8 as the expansion factor because it equals 2 cubed, and the coefficients are too small originally. In my experience, these denestings are almost always well known Pell applications.
The first expression you wrote looks like the solution to a 3rd degree using Cardano's formula. I usually have trouble getting from there towards the original expression. But I will watch the rest of the video to see if that helps
My take: x^3 - 3x - 2√5 = 0. Use x = y√5. Then 5y^3 √5 - 3y √5 - 2√5 = √5(5y^3 - 3y - 2) = 0. The sum of the coefficients is 0. So, 1 is a root. y = 1 => x = √5
I would request you to look into the matter that whenever you were writing in the lower portion of the board the writings were being covered by your speech though it was seen when it scrolled upward. Pl. consider my problem. With thanks and I appreciate you for the efforts.
let x =(√5 +2)^(1/3) + (√5 -2)^(1/3) so x*x*x = 2√5 +3x x= √5 being a solution to this one this can be rewritten as (x-√5)(x*x + √5*x + 2) = 0 or (√5 +2)^(1/3) + (√5 -2)^(1/3)= √5 Similarly let y = (√5 +2)^(1/3) - (√5 -2)^(1/3) so y*y*y = 2*2 - 3y y= 1 being a solution to this one this can be rewritten as (y-1)(y*y + y + 4) = 0 or (√5 +2)^(1/3) - (√5 -2)^(1/3)= 2 Therefore (√5 +2)^(1/3) = (√5 +1)/2
Great video! I've got some problem to solve but I don't know if my solution is correct or not, perhaps you can solve and I check Prove: sin(x)^8+cos(x)^8 >= 0.125
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
I note that the conjugate of sqrt(5)+2 is 2-sqrt(5), not sqrt(5)-2, because the radical changes when you conjugate. My idea for solving this is to add the conjugates together: (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3). This looks like the solution of a cubic equation; in fact, if you "antisolve" the expression (a+sqrt(b))^(1/3) + (a-sqrt(b))^(1/3), by comparing a and b with -q/2 and q^2/4+p^3/27, you get the equation x^3-3*(a^2-b)^(1/3)* x-2a=0. Substituting 2 and 5 for a and b gives x^3+3x-4=0. Lo and behold, 1 solves this equation. The other two roots are complex and can't be right since everything here is real. Therefore (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3) = 1. Now multiply them. (2+sqrt(5))^(1/3) * (2-sqrt(5))^(1/3) = (4-5)^(1/3) = -1. So (2+sqrt(5))^(1/3) and its conjugate satisfy x^2-x-1 = 0, from which you get phi and -1/phi as roots. So (2+sqrt(5))^(1/3) = phi.
That's pretty!
Here is the general method I use when trying to denest a cube root radical with a square root inside.
Begin by assuming the radical denests as
eq1: (A + sqrt(B))^(1/3) = X + sqrt(Y)
which if it does hold, then it can be shown (but I will not do so here) that
eq2: (A - sqrt(B))^(1/3) = X - sqrt(Y)
Now, define the product of the left hand sides of eq1 and eq2 as R
R = (A^2 - B)^(1/3)
and note that R must be a rational number for the cube root to denest. That is a necessary but not sufficient condition for denesting to succeed. Next, multiply eq1 and eq2 to get
eq3: X^2 - Y = R
which will be used later. Now add eq1 and eq2 to get
2*X = (A + sqrt(B))^(1/3) + (A - sqrt(B))^(1/3)
and cube that to get
8*X^3 = A + sqrt(B) + A - sqrt(B) + 3*R*((A + sqrt(B))^(1/3) + (A - sqrt(B))^(1/3)) = 2*A + 3*R*(2*X)
which can be rearranged to standard form for a depressed cubic equation
eq4: 4*X^3 - 3*R*X - A = 0
which, if you try to solve using Cardano's formula you will just get more nested radicals. So use the rational root theorem to find a root X1:
(any factor of A) / (± any factor of 4)
There are 6 possibilities for the denominator (±1, ±2, ±4) and N number of possibilities for the numerator (do not double for plus/minus for the numerator or you double count) so at most 6*N for the total number of possible rational roots to try in eq4 (usually fewer depending on the actual factors). Of course you can stop once you find one that works. Then you can determine Y from eq3
Y = X^2 - R
In this case, A = 2 , B = 5, and R = -1 so the cubic is
4*X^3 + 3*X - 2 = 0
so there are at most 6*2 = 12 possibilities for a rational root as (1,2) / (±1, ±2, ±4) but there are actually only 8 possibilities in this case and I find that X = 1/2 is a root, and then Y = X^2 - R = 1/4 - (-1) = 5/4 so we have
X + sqrt(Y) = 1/2 + sqrt(5/4) = (1 + sqrt(5)) / 2
Nice!
@Afnaan2701 All cubic formulas such as Cardano's formula will just give you a nested radical, which is what we already are struggling with. That is why you need to use the rational root theorem instead of a formula (or the method used to derive the formula, which is of course the same as using the formula)
@Afnaan2701 No, that is incorrect. As I said, there are formulas for the solution to cubic equations, such as Cardano's formula. The issue is that the formula contains nested radicals, so the formulas are generally not helpful in denesting nested radicals.
@Afnaan2701 I already answered your question. Twice. I am going to stop replying now.
Or if you know that phi = (root(5)+1)/2, you can note that root(5) + 2 = root(5) + 1 + 1 = 2 phi +1, and use phi^2 = phi +1 to see that phi^3 = phi^2 + phi = 2 phi +1, so cuberoot(2 phi +1) = cuberoot(phi^3) = phi.
Pretty good!
Here's a general method, with no "magic" or "deus ex-machina" in it, although the final step does involve an obvious guess. Since radicals and integers are independent bases, we know that ∛(√5+2) will be of the form a + b√5. Cubing (a + b√5), we get a^3 + 3a^2*b√5 + 15a*b^2 + 5√5*b^3. Equating coefficients of the integers of that and (√5+2), we get a^3 + 15a*b^2 = 2. Equating coefficients of the √5 parts gives 3a^2*b + 5b^3 = 1. Now, those look rather difficult to solve for a and b, but you can see a solution more easily by substituting c=2a and d=2b, then multiplying both sides of both equations by 8. That gives 3c^2*d + 5d^3 = 8 and c^3 + 15c*d^2 = 16. Looking at the coefficients indicates c=1, d=1. So a=1/2, b=1/2 and therefore ∛(√5+2) = 1/2 + √5/2.
u + 1/u = √5 provides 2 solutions u1 and u2 since u1 is = 1/u2 and viceversa. Both are positive and one is greater than 1. The other, since it's the reciprocal of tht one, is less than 1. We need the greater one. So, pick the + version
Great way of solving this radical
Loved the first way!
Glad you liked it!
@@SyberMath why would anyone ever think pf adding the cube root pf radocal 5 minus 2? I don't see why amyone would think of that ? Don't you agree?
@@leif1075 Thanks for the question! Someone who dealt with radical expressions for while and solved a good variety of problems on radicals should know how to use conjugates in certain situations. It takes practice and willingness to learn these. I know it's not always very straightforward and that's why I presented an alternative solution method.
(Root 5 plus 1 over 2): You know who I am! Say my name!
SyberMath: No.
Root 5 + 2 = 2 phi + 1
2 phi + 1 = phi+phi+1
Phi+phi+1 = phi +(phi^2)
Phi + (phi^2) = phi (phi +1)
Phi (phi +1) = phi (phi^2)
Phi (phi^2) = phi^3
Cube root of phi cubed is phi.
Who is phi? 😂😂😂
@@SyberMath Coukdbt you just cube the whole expression and work from there..dont you think this isnwhat most people would do?
@@kurtlichtenstein2325 Nice one, indeed! In this case one must know what is phi.
@@SyberMath
Root 5 +2=2phi +1
phi^2=phi+1
root 5+2=phi+phi+1=phi+phi^2=phi^3
X=phi=root 5 +2 over 2
Per que, min 9:37 , podem fer b=ka?
El sistema dius qué és homogeni, però té termes constants (eq1=2 i eq2=1)?
Merci
( escrit en Català)
With a little bit of Pell Equation theory one can just write:
√5 + 2 = ⅛⋅(5√5 + 15 + 3√5 + 1)
= ⅛⋅(5√5 + 3⋅5 + 3⋅√5 + 1)
= ⅛⋅(√5³ + 3⋅√5² + 3⋅√5 + 1)
= (½)³⋅(√5 + 1)³.
This may look magical, but is really just recognizing the Pell Eqn consequences of the second method;
and applying them.
One knows to use 8 as the expansion factor because it equals 2 cubed, and the coefficients are too small originally.
In my experience, these denestings are almost always well known Pell applications.
Nice!
Amazing, it looks like a number impossible to simplify but indeed it is! Math is magic!
Exactly!
@@SyberMath q1
Wow! Thanks again for your amazing solution and explanation, professor!
Here's how I know how to solve it: First, let α = 3√ (√5 +2), β = 3 √ (√ 5-2), then α ^3-β ^3 = 4, and (α^3)(β^3)=(αβ)^3=1, αβ=1 (as α>β>0), (α-β)^3=α^3+β^3-3αβ(α-β)=4-3(α-β), let α-β=γ, γ^3+3γ-4=0, γ^3+3γ-4=(γ-1)(γ^2+γ+4), γ=1 (as γ^2 +γ+4=0 is D=1^2-4*1*4=-15β>0 ,β= (-1 + √5) / 2, α = 3 √ (√5 + 2) = (1 + √5) / 2.
I expanded (1+√5)^3 as a trial.
I found the solution immediately.
Thanks for the Golden video!🤣 🏆
No problem
@@SyberMath !🤣 🏆
The first expression you wrote looks like the solution to a 3rd degree using Cardano's formula. I usually have trouble getting from there towards the original expression. But I will watch the rest of the video to see if that helps
f(x) = x^3 - 3x - 2v5. It's derivative is 3x^2 - 3, which has 2 real solutions, 1 and - 1. Plugging them into f(x), I get 1-3-2√5, which is
My take: x^3 - 3x - 2√5 = 0. Use x = y√5. Then 5y^3 √5 - 3y √5 - 2√5 = √5(5y^3 - 3y - 2) = 0. The sum of the coefficients is 0. So, 1 is a root. y = 1 => x = √5
Amaizing sir!!!😃😃
Thanks for all.
You are most welcome! 😊
I would request you to look into the matter that whenever you were writing in the lower portion of the board the writings were being covered by your speech though it was seen when it scrolled upward. Pl. consider my problem. With thanks and I appreciate you for the efforts.
Good point. I'll try to be more careful
wow I did not see that at all!!!
I prefer the first one.
let x =(√5 +2)^(1/3) + (√5 -2)^(1/3) so x*x*x = 2√5 +3x
x= √5 being a solution to this one
this can be rewritten as
(x-√5)(x*x + √5*x + 2) = 0
or (√5 +2)^(1/3) + (√5 -2)^(1/3)= √5
Similarly
let y = (√5 +2)^(1/3) - (√5 -2)^(1/3)
so y*y*y = 2*2 - 3y
y= 1 being a solution to this one
this can be rewritten as
(y-1)(y*y + y + 4) = 0
or (√5 +2)^(1/3) - (√5 -2)^(1/3)= 2
Therefore
(√5 +2)^(1/3) = (√5 +1)/2
2nd method is systematic but getting a and b values is lengthy
I agree
I loved both methods
Thanks for the feedback!
HOW DID YOU KNOW THE ROOT OF CUBIC WAS ROOT 5 ?? IN THIS VIDEO ?? ALSO OTHER THAN TRIAL AND ERROR IS THERE A GOOD METHOD TO FIND ROOTS OF CUBIC
There is s cubic formula.
@@oenrn too big to apply boss , especially in jee
(2plus minas 5^1/2)^1/3ans
Great video! I've got some problem to solve but I don't know if my solution is correct or not, perhaps you can solve and I check
Prove: sin(x)^8+cos(x)^8 >= 0.125
Let f : R -> R, f(x) = sin(x)^8 + cos(x)^8. Hence f'(x) = 8·sin(x)^7·cos(x) - 8·cos(x)^7·sin(x) = 8·sin(x)·cos(x)·[sin(x)^6 - cos(x)^6] = 4·sin(2·x)·[sin(x)^6 - cos(x)^6] = 0. Therefore, sin(2·x) = 0, or sin(x)^6 - cos(x)^6 = 0.
If sin(2·x) = 0, then 2·x = n·π, with n being an element of Z. Thus x = n·π/2, and f(n·π/2) = sin(n·π/2)^8 + cos(n·π/2)^8 = a(n) + b(n), where a(n) = 0 if n is even, a(n) = 1 if n is odd, and b(n) = 1 if n is even, b(n) if is odd. Hence a(n) + b(n) = 1 for all n, and f(n·π/2) = 1 for all n.
If sin(x)^6 - cos(x)^6 = 0, then [sin(x)^3 + cos(x)^3]·[sin(x)^3 - cos(x)^3] = 0. Therefore, sin(x)^3 = cos(x)^3, or sin(x)^3 = -cos(x)^3. Thus tan(x)^3 = 1, or tan(x)^3 = -1, so tan(x) = -1 or tan(x) = 1. Therefore, x = n·π + π/4, or x = n·π - π/4.
f(n·π + π/4) = sin(n·π + π/4)^8 + cos(n·π + π/4)^8 = [sin(n·π)·cos(π/4) + cos(n·π)·sin(π/4)]^8 + [cos(n·π)·cos(π/4) - sin(n·π)·sin(π/4)] = cos(n·π)^8·sin(π/4)^8 + cos(n·π)^8·cos(π/4)^8 = sin(π/4)^8 + cos(π/4)^8 = f(π/4), while f(n·π - π/4) = sin(n·π - π/4)^8 + cos(n·π - π/4)^8 = [sin(n·π)·cos(π/4) - cos(n·π)·sin(π/4)]^8 + [cos(n·π)·cos(π/4) + sin(n·π)·sin(π/4)]^8 = cos(n·π)^8·sin(π/4)^8 + cos(n·π)^8·cos(π/4)^8 = sin(π/4)^8 + cos(π/4)^8 = f(π/4).
f(π/4) = sin(π/4)^8 + cos(π/4)^8 = [1/sqrt(2)]^8 + [1/sqrt(2)]^8 = 1/2^4 + 1/2^4 = 2/2^4 = 1/2^3 = 1/8 = 0.125.
Q. E. D.
@@angelmendez-rivera351 Oh, man! You are a legend!
@@SyberMath Thank you!
@@angelmendez-rivera351 Absolutely!
@@angelmendez-rivera351 damn you are a beast, but the problem is I can't use derivatives
Tx pal.
You're welcome!
OMG just grab the calculator 😂😂
or the computer! 😁
u+1/ugreater than equal to 2
cool!
a=b=2
First is better !!!!!!!!
Thanks for the feedback, Tony!
1st method is better
2nd method is easier.
Or you can use a
Calculator and solve this problem in 10 seconds🤠
That's cheating! 😂