Two solutions: A) recall the algebraic identity a^3 + b^3 + c^3 = 3abc if a+b+c=0. Now a and b are the two roots on the left side and you see easily that a*b= -1 and a+b=x. So a+b - x = 0. Now plug in c=-x in above equation leading to: x^3 + 3x -14= 0. Factorizing gives x=2 and two other unvalid complex roots because x must be rational. The solution of the question is x=2 B) Set a + b = 1 + z root (f) and cube both sides , then compare coefficients yielding a= 1+root(2). Since a*b=-1 we get b= 1-root(2). So the answer again is a+b=2
The first radical denests to 1 + sqrt(2) and the second radical denests to 1 - sqrt(2), so the sum is 2. I have posted methods for denesting radicals in previous videos, so I did not include the details in this comment.
In your first method you are using the following theorem: If a, b, c, d are rational numbers and √b and √d are irrational then a + √b = c + √d implies a = c and b = d This is often taken for granted when denesting nested roots, but it *is* a theorem, therefore it should be proved. Clearly, this does not apply if √b and √d are rational (take a = 2, b = 4, c = 3, d = 1, then a+√b = c+√d but a ≠ c and b ≠ d). The proof of this theorem is straightforward: a + √b = c + √d implies √b = (c − a) + √d Squaring both sides gives b = (c − a)² + d + 2(c − a)√d and subtracting (c−a)²+d from both sides then gives (b − d) − (c − a)² = 2(c − a)√d Now the left hand side is rational, therefore the right hand side must also be rational. But √d is irrational, therefore the right hand side can only be rational if c − a = 0 which implies a = c. With c−a = 0 the equation (b−d)−(c−a)² = 2(c−a)√d becomes b − d = 0 which implies b = d and this completes the proof.
why you do this? My proof is simpler: The first equation becomes a - c = root(d) - root(b). The left side is rational and the right side is irrarional. or n/m cannot be equal to a irrational number because it is rational.
@@hans-rudigerdrzimmermann This is not a correct proof because the difference of two irrational numbers need not be irrational. You are *assuming* that the difference √d − √b can not be rational unless d = b, but that is the very thing you need to prove.
@@SyberMath Hey, are u a prophet? I see the exact question on an entrance test this year. Edit: Good luck for all of the students that watch this before taking this exam.
OK. Challenge accepted: Want: nthrt(a+bi) Notice, we generally use the Cartesian system, in which every point is uniquely identified by the distance from two reference lines, called the x and y axis. But we can also define a unique point by the distance from a reference *point*, and the direction in which we are looking (if we know both). So let's try that. Using some trigonometry, we can figure out the distance from 0 (take the origin as 0 just like the Cartesian system). Since the a, b (from a+bi) and the distance from the origin (let's call it r) form a right triangle (look at a picture for better understanding), we can say that r²=a²+b² and r=√(a²+b²). We take the positive sqrt because r is the distance so it's always positive. Now we are done with the distance. Let's find the angle now. Using trigonometry, we find that the tangent of the angle (let's call it θ) is b/a. Let's take θ=arctan(b/a) Now we have one more task. To figure out a way to convert a+bi to its rotational (Polar as we call it) form. If we take cos(θ), we get a/r. Therefore, a=rcosθ. Similarly, sinθ =b/r. Therefore, b=rsinθ. Therefore, a+bi=rcosθ+irsinθ. Factoring out r, a+bi=r(cosθ+isinθ) Using Euler's formula, we get re^iθ.
Part 2:- Now, we want the nthrt(a+bi). With our new-found knowledge, we can say that a+bi=re^iθ, where r=√(a²+b²), and θ=arctan(b/a). Therefore, (a+bi)^1/n is the same as re^(iθ/n). Problem Solved!!!!!
Part 3:- Let's take some examples. A nice one would be sqrt(i). I is the same as 0+1i, and therefore r=1 and θ=arctan(1/0), which if you take the limit is π/2. Therefore, i=e^(iπ/2), and √i=e^(iπ/4). You can expand that using Euler's formula, cos(π/4) is 1/√2 and sin(π/4) is 1/√2, and you have the answer: 1/√2+1/√2i !!!!!!!! If you are looking for the other root, it is the negative of the root you found. Therefore, for the equation z²=i, z=±(1/√2+1/√2i) #YAY!!!!
Writing a for (7 + 5√2) ^(1/3) and b for (7 -5√2) ^(1/3) one gets a^3 + b^3 = 14 and ab = -1 Hereby (a+b)^3 = a^3 + b^3 + 3ab *(a+b) Hereby a+b is a solution of x^3 = 14 - 3x or x^3 + 3x -14 = 0 or (x+2)(x^2 - 2x + 7) = 0 Hereby x = -2, 1 + i*√6, 1- i*√6
Your factorisation is incorrect. x³ + 3x − 14 = (x − 2)(x² + 2x + 7) The zeros of the cubic are x₁ = 2, x₂ = −1 + i√6, x₃ = −1 − i√6. Setting a = ∛(7 + 5√2) = 1 + √2, b = ∛(7 − 5√2) = 1 − √2 these zeros can be expressed as x₁ = a + b x₂ = ε₁a + ε₂b x₃ = ε₂a + ε₁b where ε₁ = −½ + i·½√3 ε₂ = −½ − i·½√3 are the complex cube roots of unity, so we have x₁ = a + b x₂ = −½(a + b) + i·½√3·(a − b) x₃ = −½(a + b) − i·½√3·(a − b) As expected, this implies x₁ + x₂ + x₃ = 0 x₁x₂ + x₁x₃ + x₂x₃ = −3ab x₁x₂x₃ = a³ + b³
That's a matter of opinion and personal experience. In fact this is pretty basic conventional algebra. Things like denesting nested radicals were part of the standard highschool curriculum in my country a century ago, but this is no longer taught in highschools.
This is simple algebra which you can check for yourself using the identities (a+b)³=a³+3a²b+3ab²+b³ and (a−b)³=a³−3a²b+3ab²−b³ but I'll explain it anyway. Let's assume we have a nested cube root ∛(p + √q) where p and q are *rational* and √q is *irrational* and let's also assume that this nested cube root can be denested, that is, there exist rational numbers a and b with √b irrational such that ∛(p + √q) = a + √b This means that the cube of a + √b is equal to p + √q, so (a + √b)³ = p + √q Expanding the left hand side gives us a³ + 3a²√b + 3ab² + b²√b = p + √q which can be rearranged as (a³ + 3ab²) + (3a² + b²)√b = p + √q Now, it's important to remember that a, b, p and q are all assumed to be *rational* numbers and that √b and √q are assumed to be irrational numbers. So, (a³ + 3ab²) and (3a² + b²) are both rational. But the product of a rational number and an irrational number is irrational, so (3a² + b²)√b is irrational. We can write this as a single square root, since (3a² + b²) is the square root of (3a² + b²)² so we can write this as √((3a² + b²)²b) and this is therefore an irrational square root of the rational number (3a² + b²)²b. We can now rewrite (a³ + 3ab²) + (3a² + b²)√b = p + √q as (a³ + 3ab²) + √((3a² + b²)²b) = p + √q There is an important theorem which I discuss and prove in another comment on this video and which says that if a,b,c,d are rational numbers and √b and √d are irrational then a+√b=c+√d implies both a=c and b=d. This is usually taken for granted, but it *is* a theorem which can (and therefore must) be proved to justify its use. If we apply this theorem to (a³ + 3ab²) + √((3a² + b²)²b) = p + √q we can conclude that we have both a³ + 3ab² = p and (3a² + b²)²b = q Now, let's take a look at the conjugate of a + √b which is a − √b. If we cube this, we get (a − √b)³ = a³ − 3a²√b + 3ab² − b²√b which can be rearranged to give (a − √b)³ = (a³ + 3ab²) − (3a² + b²)√b or (a − √b)³ = (a³ + 3ab²) − √((3a² + b²)²b) But since a³ + 3ab² = p and (3a² + b²)²b = q this means that we have (a − √b)³ = p − √q and therefore ∛(p − √q) = a − √b So, we have proved the following *Theorem* If p and q are rational numbers and √q is irrational and if the nested cube root ∛(p + √q) can be denested as ∛(p + √q) = a + √b where a and b are rational numbers and √b is irrational, then the nested cube root ∛(p − √q) can be denested as ∛(p − √q) = a − √b which answers your question. Note that this theorem also applies to a nested cube root such as ∛(7+5√2) since we can write this as ∛(7+√50). So, if ∛(7+5√2) = a+b√2 = a+√(2b²) where a and b are rational whereas b√2 = √(2b²) is irrational, then the above theorem about denestable cube roots guarantees that ∛(7−5√2) = ∛(7−√50) is also denestable and that we will then have ∛(7−5√2) = ∛(7−√50) = a−√(2b²) = a−b√2 and consequently ∛(7+5√2) + ∛(7−5√2) = 2a.
Suppose ∛(7+5√2)=a+b√2... Cubing both sides and arranging like terms we get 7+5√2=(a+b√2)³ 7+5√2=a³ + 6ab²+(2b³+3a²b)√2 and this gives us a³ + 6ab²=7 and 2b³+3a²b=5 Let's now cube a-b√2 (a-b√2)³=a³ + 6ab²-(2b³+3a²b)√2=7-5√2 ∛(7-5√2)=a-b√2 Therefore, ∛(7+5√2) and ∛(7-5√2) are conjugates...
Actually this is what Cardano's formula gives you as the real root of the equation x³ + 3x − 14 = 0 and this is also the equation obtained in this video with the second method, which amounts to reversing the usual solution method of a reduced cubic equation and working your way back from the solution expressed as a sum (or: difference) of two nested cube roots to the cubic equation which has the given expression as one of its solutions.
Two solutions:
A) recall the algebraic identity a^3 + b^3 + c^3 = 3abc if a+b+c=0. Now a and b are the two roots on the left side and you see easily that a*b= -1 and a+b=x.
So a+b - x = 0. Now plug in c=-x in above equation leading to: x^3 + 3x -14= 0. Factorizing gives x=2 and two other unvalid complex roots because x must be rational. The solution of the question is x=2
B) Set a + b = 1 + z root (f) and cube both sides , then compare coefficients yielding a= 1+root(2). Since a*b=-1 we get b= 1-root(2). So the answer again is a+b=2
I solved on my own and it exactly matched with 2nd method. All credits go to U. May GOD bless U
The first radical denests to 1 + sqrt(2) and the second radical denests to 1 - sqrt(2), so the sum is 2.
I have posted methods for denesting radicals in previous videos, so I did not include the details in this comment.
Yes, that's how I did it too. By cubing both sides, you can prove those equalities.
Good, look at my other more direct solution in the commentary
I'm interested in that de-nesting comments so, would you so kind posting links to those comments?
"Denesting radicals" . . . sounds like a Russian war maneuver.
In your first method you are using the following theorem:
If a, b, c, d are rational numbers and √b and √d are irrational then a + √b = c + √d implies a = c and b = d
This is often taken for granted when denesting nested roots, but it *is* a theorem, therefore it should be proved. Clearly, this does not apply if √b and √d are rational (take a = 2, b = 4, c = 3, d = 1, then a+√b = c+√d but a ≠ c and b ≠ d). The proof of this theorem is straightforward:
a + √b = c + √d
implies
√b = (c − a) + √d
Squaring both sides gives
b = (c − a)² + d + 2(c − a)√d
and subtracting (c−a)²+d from both sides then gives
(b − d) − (c − a)² = 2(c − a)√d
Now the left hand side is rational, therefore the right hand side must also be rational. But √d is irrational, therefore the right hand side can only be rational if c − a = 0 which implies a = c. With c−a = 0 the equation (b−d)−(c−a)² = 2(c−a)√d becomes b − d = 0 which implies b = d and this completes the proof.
why you do this? My proof is simpler: The first equation becomes a - c = root(d) - root(b). The left side is rational and the right side is irrarional. or n/m cannot be equal to a irrational number because it is rational.
@@hans-rudigerdrzimmermann This is not a correct proof because the difference of two irrational numbers need not be irrational. You are *assuming* that the difference √d − √b can not be rational unless d = b, but that is the very thing you need to prove.
Very good!
@@SyberMath Hey, are u a prophet?
I see the exact question on an entrance test this year.
Edit: Good luck for all of the students that watch this before taking this exam.
Thanks 🙏
Very elegant
Thank you!
Have you try to simplify a "n" root of a imaginary number? Since sqrt i = {sqrt2} / 2 + I * {sqrt 2} / 2, it's possible
This is easy to do if you rewrite the complex number in polar form, but you generally can't do it algebraically for n > 2.
OK. Challenge accepted:
Want: nthrt(a+bi)
Notice, we generally use the Cartesian system, in which every point is uniquely identified by the distance from two reference lines, called the x and y axis. But we can also define a unique point by the distance from a reference *point*, and the direction in which we are looking (if we know both). So let's try that. Using some trigonometry, we can figure out the distance from 0 (take the origin as 0 just like the Cartesian system). Since the a, b (from a+bi) and the distance from the origin (let's call it r) form a right triangle (look at a picture for better understanding), we can say that r²=a²+b² and r=√(a²+b²). We take the positive sqrt because r is the distance so it's always positive. Now we are done with the distance. Let's find the angle now.
Using trigonometry, we find that the tangent of the angle (let's call it θ) is b/a. Let's take θ=arctan(b/a)
Now we have one more task. To figure out a way to convert a+bi to its rotational (Polar as we call it) form. If we take cos(θ), we get a/r. Therefore, a=rcosθ. Similarly, sinθ =b/r. Therefore, b=rsinθ. Therefore,
a+bi=rcosθ+irsinθ. Factoring out r,
a+bi=r(cosθ+isinθ) Using Euler's formula, we get re^iθ.
n th root of (a+ib) = (a+ib)^(1/n)
= (r(e^(i theta))^(1/n))
= r(e^(i theta/n)), where
r = sqrt(a^2 + b^2) and theta = tan inverse (b/a)
Part 2:-
Now, we want the nthrt(a+bi). With our new-found knowledge, we can say that a+bi=re^iθ, where r=√(a²+b²), and θ=arctan(b/a). Therefore, (a+bi)^1/n
is the same as re^(iθ/n). Problem Solved!!!!!
Part 3:-
Let's take some examples. A nice one would be sqrt(i). I is the same as 0+1i, and therefore r=1 and θ=arctan(1/0), which if you take the limit is π/2. Therefore, i=e^(iπ/2), and √i=e^(iπ/4). You can expand that using Euler's formula, cos(π/4) is 1/√2 and sin(π/4) is 1/√2, and you have the answer:
1/√2+1/√2i !!!!!!!!
If you are looking for the other root, it is the negative of the root you found. Therefore, for the equation z²=i,
z=±(1/√2+1/√2i)
#YAY!!!!
Writing a for (7 + 5√2) ^(1/3)
and b for (7 -5√2) ^(1/3) one gets
a^3 + b^3 = 14 and ab = -1
Hereby (a+b)^3
= a^3 + b^3 + 3ab *(a+b)
Hereby a+b is a solution of
x^3 = 14 - 3x
or x^3 + 3x -14 = 0
or (x+2)(x^2 - 2x + 7) = 0
Hereby x = -2, 1 + i*√6, 1- i*√6
Your factorisation is incorrect.
x³ + 3x − 14 = (x − 2)(x² + 2x + 7)
The zeros of the cubic are x₁ = 2, x₂ = −1 + i√6, x₃ = −1 − i√6. Setting
a = ∛(7 + 5√2) = 1 + √2,
b = ∛(7 − 5√2) = 1 − √2
these zeros can be expressed as
x₁ = a + b
x₂ = ε₁a + ε₂b
x₃ = ε₂a + ε₁b
where
ε₁ = −½ + i·½√3
ε₂ = −½ − i·½√3
are the complex cube roots of unity, so we have
x₁ = a + b
x₂ = −½(a + b) + i·½√3·(a − b)
x₃ = −½(a + b) − i·½√3·(a − b)
As expected, this implies
x₁ + x₂ + x₃ = 0
x₁x₂ + x₁x₃ + x₂x₃ = −3ab
x₁x₂x₃ = a³ + b³
This is the Cardano solution to x(x²+3)=14.
Mind-boggling.
That's a matter of opinion and personal experience. In fact this is pretty basic conventional algebra. Things like denesting nested radicals were part of the standard highschool curriculum in my country a century ago, but this is no longer taught in highschools.
Please explain why these 2 numbers are conjugate. Adding your numbers make it equal to 2a.
This is simple algebra which you can check for yourself using the identities (a+b)³=a³+3a²b+3ab²+b³ and (a−b)³=a³−3a²b+3ab²−b³ but I'll explain it anyway.
Let's assume we have a nested cube root
∛(p + √q)
where p and q are *rational* and √q is *irrational* and let's also assume that this nested cube root can be denested, that is, there exist rational numbers a and b with √b irrational such that
∛(p + √q) = a + √b
This means that the cube of a + √b is equal to p + √q, so
(a + √b)³ = p + √q
Expanding the left hand side gives us
a³ + 3a²√b + 3ab² + b²√b = p + √q
which can be rearranged as
(a³ + 3ab²) + (3a² + b²)√b = p + √q
Now, it's important to remember that a, b, p and q are all assumed to be *rational* numbers and that √b and √q are assumed to be irrational numbers. So, (a³ + 3ab²) and (3a² + b²) are both rational. But the product of a rational number and an irrational number is irrational, so
(3a² + b²)√b
is irrational. We can write this as a single square root, since (3a² + b²) is the square root of (3a² + b²)² so we can write this as
√((3a² + b²)²b)
and this is therefore an irrational square root of the rational number (3a² + b²)²b. We can now rewrite
(a³ + 3ab²) + (3a² + b²)√b = p + √q
as
(a³ + 3ab²) + √((3a² + b²)²b) = p + √q
There is an important theorem which I discuss and prove in another comment on this video and which says that if a,b,c,d are rational numbers and √b and √d are irrational then a+√b=c+√d implies both a=c and b=d. This is usually taken for granted, but it *is* a theorem which can (and therefore must) be proved to justify its use. If we apply this theorem to
(a³ + 3ab²) + √((3a² + b²)²b) = p + √q
we can conclude that we have both
a³ + 3ab² = p
and
(3a² + b²)²b = q
Now, let's take a look at the conjugate of a + √b which is a − √b. If we cube this, we get
(a − √b)³ = a³ − 3a²√b + 3ab² − b²√b
which can be rearranged to give
(a − √b)³ = (a³ + 3ab²) − (3a² + b²)√b
or
(a − √b)³ = (a³ + 3ab²) − √((3a² + b²)²b)
But since a³ + 3ab² = p and (3a² + b²)²b = q this means that we have
(a − √b)³ = p − √q
and therefore
∛(p − √q) = a − √b
So, we have proved the following
*Theorem*
If p and q are rational numbers and √q is irrational and if the nested cube root ∛(p + √q) can be denested as
∛(p + √q) = a + √b
where a and b are rational numbers and √b is irrational, then the nested cube root ∛(p − √q) can be denested as
∛(p − √q) = a − √b
which answers your question. Note that this theorem also applies to a nested cube root such as ∛(7+5√2) since we can write this as ∛(7+√50). So, if ∛(7+5√2) = a+b√2 = a+√(2b²) where a and b are rational whereas b√2 = √(2b²) is irrational, then the above theorem about denestable cube roots guarantees that ∛(7−5√2) = ∛(7−√50) is also denestable and that we will then have ∛(7−5√2) = ∛(7−√50) = a−√(2b²) = a−b√2 and consequently ∛(7+5√2) + ∛(7−5√2) = 2a.
Suppose ∛(7+5√2)=a+b√2...
Cubing both sides and arranging like terms we get
7+5√2=(a+b√2)³
7+5√2=a³ + 6ab²+(2b³+3a²b)√2 and this gives us
a³ + 6ab²=7 and 2b³+3a²b=5
Let's now cube a-b√2
(a-b√2)³=a³ + 6ab²-(2b³+3a²b)√2=7-5√2
∛(7-5√2)=a-b√2
Therefore, ∛(7+5√2) and ∛(7-5√2) are conjugates...
This actually looks like Cardano’s Cubic Formula!
Actually this is what Cardano's formula gives you as the real root of the equation
x³ + 3x − 14 = 0
and this is also the equation obtained in this video with the second method, which amounts to reversing the usual solution method of a reduced cubic equation and working your way back from the solution expressed as a sum (or: difference) of two nested cube roots to the cubic equation which has the given expression as one of its solutions.
The input looks like a result of the cubic formula
it is
Wow this question has just got in the test to a gifted high school in Vietnam this morning
😲😁🤩
way to go!
Yesss!
The Master has done it once again!
Thank you for the kind words, Moe! 🤗🥰
I don't know more adjectives to post here.
I made it for the 2nd method, but the 1st one is really more interesting.
🇧🇷🇧🇷🇧🇷🇧🇷Brazil🇧🇷🇧🇷🇧🇷🇧🇷
Greetings!
Thanks
You're welcome!
great question
Thanks
x = (7 + 5 √2)^(1/3) + (7 - 5 √2)^(1/3)
x = a + b
a^3 = 7 + 5 √2
b^3 = 7 - 5 √2
a^3 + b^3 = 14
a.b = (7)^2-(5 √2)^2 = -1
(a+b)^3 = a^3 + b^3 + 3 ab(a+b)
(a+b)^3 = 14 + 3 * (-1) (a+b)
x^3 = 14 - 3 x
x^3 + 3 x - 14 = 0
(x-2)(x^2+2 x+7) = 0
x=2
(x-2)(x²+2x+7)
Ans =2
x^3+3x-14=0
I assigned you the father of Algebra
Aww, thank you!!! 🤗🥰
好，奖 1 万元
I also got x=2 as the only real solution!
Me so