A nice Algebra Challenge | Math Olympiad Simplification | You Should know this trick!
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- เผยแพร่เมื่อ 14 พ.ค. 2024
- A nice Algebra Challenge | Math Olympiad Simplification | You Should know this trick! # radical#olympiad #simplification#exam
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Suppose -2+√5 = (u +√v)³ where u & v are rational & v is not the square of a rational.
(u² - v)³ = (u + √v)³.(u - √v)³ = (- 2 + √5).(- 2 - √5) = 4 - 5 = - 1 = (-1)³, so u and v exists.
If we did not get the cube of a rational number, then u and v would not have existed.
Now - 2 + √5 = (u³ + 3uv) + (3u² + v)√v. So - 2 = (u³ + 3uv) and √5 = (3u² + v)√v.
Therefore (u-√v)³ = (u³+3uv) - (3u²+v)√v = - 2 - √5. So - 1 = 4 - 5 = (- 2 +√5).(- 2 -√5)
= [(u + √v)³].[(u - √v)³] = (u² - v)³. Hence (u² - v) = ∛(-1) = - 1 & so v = u² - (-1) = u² +1.
Therefore - 2 = (u³ + 3u.v) = u³ + 3u.(u² + 1) = 4.u³ + 3u. So 4.u³ + 3.u + 2 = 0. By the
Rational Root theorem, u = -½, bec. 4u³ + 3u + 2 = (2u+1).(2u² - u +2) & the quadratic
part has no real roots. ∴ v = u²+1 = ⁵/₄ . So ∛(-2 +√5) = (u+√v) = -½ + √(⁵/₄) = (√5 -1)/2.
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Or one can write sqrt(5) - 2 as 1/8 x (8 sqrt(5) - 16) and discover that (8 sqrt(5) - 16) is a perfect cube using the (a - b)^3 formula
There is a standard procedure for denesting denestable cube roots like ∛(√5 − 2) which is more efficient than the procedure shown in the video and which does not require solving a quadratic equation. First, rewrite ∛(√5 − 2) in standard form as ∛(2 − √5) and assume there exist _rational_ numbers x and y with √y _irrational_ such that
(1) ∛(2 + √5) = x + √y
then we must also have
(2) ∛(2 − √5) = x − √y
From (1) and (2) we have x² − y = (x + √y)(x − √y) = ∛(2 + √5)·∛(2 − √5) = ∛((2 + √5)(2 − √5)) = ∛(2² − (√5)²) = ∛(4 − 5) = ∛(−1) = −1, so
(3) x² − y = −1
Also from (1) the cube of x + √y must equal 2 + √5 so we have (x + √y)³ = x³ + 3x²√y + 3xy + y√y = (x³ + 3xy) + (3x² + y)√y = 2 + √5 which implies
(4) x³ + 3xy = 2
From (3) we have y = x² + 1 and substituting this in (4) we get
x³ + 3x(x² + 1) = 2
which gives
(5) 4x³ + 3x − 2 = 0
Since x must be rational, we are looking for rational solutions of this cubic equation. According to the rational root theorem, for any potential rational solution m/n or −m/n of (5), m must divide the absolute value of constant term, which is 2, and n must divide (the absolute value of) the coefficient of the cubic term, which is 4, so m can be 1 or 2 and n can be 1 or 2 or 4.
However, we can also see that (a) the polynomial on the left hand side of (5) is strictly increasing on the real number line and (b) the left hand side of (5) is negative for any negative x. This means, first of all, that (5) can only have a single real solution and, secondly, that any real solution and _eo ipso_ any rational solution must be positive. So, we only need to test ¼, ½, 1 and then we quickly find that x = ½ is a solution and consequently the only real solution of (5).
With x = ½ we have y = x² + 1 = (½)² + 1 = ¼ + 1 = ⁵⁄₄ and so we have
∛(2 + √5) = ½ + √(⁵⁄₄)
∛(2 − √5) = ½ − √(⁵⁄₄)
which can also be written as
∛(2 + √5) = ½ + ½√5
∛(2 − √5) = ½ − ½√5
Of course, ∛(2 − √5) = ½ − ½√5 implies that we also have
∛(√5 − 2) = ½√5 − ½ = (√5 − 1)/2
Reference: Kaidy Tan, Finding the Cube Root of Binomial Quadratic Surds. _Mathematics Magazine_ Vol. 39, No. 4 (Sep., 1966), pp. 212-214 (JSTOR 2688084)
Thanks for shedding light on this Math Olympiad problem
Now I've watched your video. There's one thing, that I've noticed with your calculations. You often ignore symmetry.
If you have a (skew-) symmetrical problem like "solve x - y = -1 and xy = 1", you know, you can simply exchange x with -y.
The solutions will also show this. The -y, that you calculate for the first solution x_1, is always the second solution for x (x_2), which has -x_1 as its y (y_2). You don't need to calculate twice. Other than that a comprehensive video!
Math Olympiad Simplification: ³√(√5 - 2) = ?
Let: ³√(√5 - 2) = a + b√5 > 0; a, b ϵR
√5 - 2 = (a + b√5)³ = a³ + 3a²b√5 + 3a(b√5)² + (b√5)³
= b(3a² + 5b²)√5 + a(a² + 15b²); b(3a² + 5b²) = 1, a(a² + 15b²) = - 2
When: a = - b; b(3b² + 5b²) = 8b³ = 1, b³ = 1/8; b = 1/2, a = - 1/2
b = - a; a(a² + 15a²) = 16a³ = - 2, a³ = - 1/8; a = - 1/2, b = 1/2; Proved
³√(√5 - 2) = a + b√5 = (√5 - 1)/2
Answer check:
[(√5 - 1)/2]³ = [(√5)³ - 3(√5)² + 3√5 - 1]/8 = (8√5 - 16)/8 = √5 - 2
Final answer:
³√(√5 - 2) = (√5 - 1)/2
Oh, this is easy, because we know, that sqrt(5) - 2 is the 3rd power of 1/ϕ (ϕ being the Golden Ratio).
So crt(ϕ^-3) = ϕ^-1 = 1/ϕ = ϕ - 1 = (sqrt(5) - 1) / 2.
Why is 1/ϕ = ϕ - 1 or 1 = ϕ^2 - ϕ. It's because ϕ and -1/ϕ are the roots of x^2 - x - 1 = 0.
On the other hand, you can define ϕ^(k+2) = ϕ^(k+1) + ϕ^k for all k € IZ.
Just testing the initial claim: ϕ^-3 = ϕ^-1 - ϕ^-2 = (ϕ - 1) - (1 - ϕ^-1) = (ϕ - 1) + (1/ϕ - 1) = (ϕ + 1/ϕ) - 2 = sqrt(5) - 2.
Before you ask: ϕ^-4 is more complicated, ϕ^-4 = 7/2 - 3/2 sqrt(5).
But it's also not so surprising, you can easily create a sequence for the coefficents!
That would be worth of a video: "If ϕ = (sqrt(5) + 1) / 2 and ϕ^k = a_k + b_k sqrt(5), what is { a_k } and { b_k } ?"
Hint: Initial values are a_0 = 1, b_0 = 0, a_1 = 1/2, b_1 = 1/2. What is the rule for a_k and b_k?
Isn't it already simple enough? 😅
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