Since the cubic term is missing, we write x^4-72x-17=(x^2+ax+17)(x^2-ax-1). Thus, a^2=16 and 18a=72. So, a=4. Thus, the given equation is tantamount to (x^2+4x+17)(x^2-4x-1)=0. If x^2+4x+17=0. x=-2+/-√13 i. If x^2-4x-1=0, x=2+/-√5.
Alternate easy approach: (1) First draw an approximate graph to get some more information. (The derivative, 4x^3 - 72 = 0, has one zero, so there is exactly one local minimum and the graph tends to +inf on both sides). The local minimum occurs at x0 = cuberoot(18) and f(x0) < 0, so it becomes clear that there are two real roots and two imaginary roots. Furthermore neither real root is an integer, they are somewhere between (-1,0) and (+4,+5), so we're not going to get away with guessing integers. (2) Since this is an "Oxford Admission Test Question" we assume the problem can be cheesed, so given the hints from (1), assume the original polynomial can be factored into two quadratics, one with real roots and the other with imaginary roots which we can solve individually with the quadratic formula. Furthermore assume that the constant terms of the quadratics are +/- 1 and +/- 17 in keeping with cheese strategy, this is a reasonable guess because 17 is prime. (3) Option 1 is f(x) = (x^2 + ax + 1)(x^2 + bx - 17) and Option 2 is f(x) = (x^2 + ax - 1)(x^2 + bx + 17) for reals {a,b}. Multiply the quadratics in both cases and note that the x^3 and x^2 coefficients of the product must be zero. It becomes clear that Option 1 is impossible because a+b=0 (from x^3 term) and ab = 16 (from x^2 term) which has no solution. Option 2 on the other hand gives a = -4, b = +4. So we have f(x) = (x^2 - 4x - 1)(x^2 + 4x + 17) = 0 which has roots 2 +/- sqrt(5) and -2 +/- sqrt(13) i.
Thanks for sharing your approach! I'm always happy to see different ways of solving problems. 🙏That's a very creative and elegant solution! 💯 I love how you used the information about the graph and the problem's context to simplify the problem. 💪🔥
Why do you look for real solutions only for "n", and accept complex solutions for "x"? The domain should be defined in the subject to avoid this debate
@@germanalexandru I didn't find it to be simple. I started with the 4x4 system you stated, which I checked and agreed with. By substitutions to isolate b I arrived at a sixth-degree (b^6) polynomial in b. Using the Rational Root Theorem I found (b+1) and (b-17) to be factors of it. The remaining fourth-degree (b^4) polynomial did not have rational roots. Owing to symmetry I could choose either -1 or +17 for b, so I chose -1. This gave d=17 and in turn a=-4 and c=4. So {a,b,c,d} = {-4,-1,4,17} is a solution. By symmetry {4,17,-4,-1} is another solution. If you have a way to solve that 4x4 without generating a horrendous polynomial, I would like to see it. Once I knew I had (x^2-4x-1)(x^2+4x+17)=0 the roots came right out: x = 2+-sqrt(5) and x = -2+-sqrt(13)i.
@brianwade4179 IT can only be +17 and -1, because the sum of them must be 4^2 and can not be - 16 or You get "i". Interesting is what if it is a valid choice. If I remember I will test it.
Good morning, dear colleague ! The general strategy is good and natural : to get a perfect square in each member from an unknown trinomial. On the other hand, factorization by (n - 8) is laborious. As always, too many details...
Since the cubic term is missing, we write x^4-72x-17=(x^2+ax+17)(x^2-ax-1). Thus, a^2=16 and 18a=72. So, a=4. Thus, the given equation is tantamount to (x^2+4x+17)(x^2-4x-1)=0. If x^2+4x+17=0. x=-2+/-√13 i. If x^2-4x-1=0, x=2+/-√5.
Interesting approach! 💯Thanks for sharing your perspective! 🙏
1. What would you have done if the constant term had not been prime?
2. Why did you pick +ax+17 and -ax-1 instead of +ax-17 and -ax+1?
Alternate easy approach:
(1) First draw an approximate graph to get some more information. (The derivative, 4x^3 - 72 = 0, has one zero, so there is exactly one local minimum and the graph tends to +inf on both sides). The local minimum occurs at x0 = cuberoot(18) and f(x0) < 0, so it becomes clear that there are two real roots and two imaginary roots. Furthermore neither real root is an integer, they are somewhere between (-1,0) and (+4,+5), so we're not going to get away with guessing integers.
(2) Since this is an "Oxford Admission Test Question" we assume the problem can be cheesed, so given the hints from (1), assume the original polynomial can be factored into two quadratics, one with real roots and the other with imaginary roots which we can solve individually with the quadratic formula. Furthermore assume that the constant terms of the quadratics are +/- 1 and +/- 17 in keeping with cheese strategy, this is a reasonable guess because 17 is prime.
(3) Option 1 is f(x) = (x^2 + ax + 1)(x^2 + bx - 17) and Option 2 is f(x) = (x^2 + ax - 1)(x^2 + bx + 17) for reals {a,b}.
Multiply the quadratics in both cases and note that the x^3 and x^2 coefficients of the product must be zero.
It becomes clear that Option 1 is impossible because a+b=0 (from x^3 term) and ab = 16 (from x^2 term) which has no solution.
Option 2 on the other hand gives a = -4, b = +4.
So we have f(x) = (x^2 - 4x - 1)(x^2 + 4x + 17) = 0 which has roots 2 +/- sqrt(5) and -2 +/- sqrt(13) i.
Thanks for sharing your approach! I'm always happy to see different ways of solving problems. 🙏That's a very creative and elegant solution! 💯 I love how you used the information about the graph and the problem's context to simplify the problem. 💪🔥
Why do you look for real solutions only for "n", and accept complex solutions for "x"? The domain should be defined in the subject to avoid this debate
Choosing the real solution of the resolvent cubic simplifies calculations.
x⁴ - 72x - 17 = 0 ← it would be interesting to have 2 squares on the left side (because power 4 and power 2)
Let's rewrite the equation by introducing a variable "λ" which, if carefully chosen, will produce 2 squares on the left side
Let's tinker a bit with x⁴ as the beginning of a square: x⁴ = (x² + λ)² - 2λx² - λ²
x⁴ - 72x - 17 = 0 → where: x⁴ = (x² + λ)² - 2λx² - λ²
(x² + λ)² - 2λx² - λ² - 72x - 17 = 0
(x² + λ)² - [2λx² + λ² + 72x + 17] = 0 → let's try to get a second member as a square
(x² + λ)² - [2λx² + 72x + (λ² + 17)] = 0 → a square into […] means that Δ = 0 → let's calculate Δ
Δ = (72)² - 4.(2λ).(λ² + 17) → then, Δ = 0
(72)² - 4.(2λ).(λ² + 17) = 0
4.(2λ).(λ² + 17) = 72²
8λ.(λ² + 17) = 72²
λ.(λ² + 17) = 72²/8
λ³ + 17λ = 648 → let's try to find the cube of a number (< 648)
1³ = 1
2³ = 8
3³ = 27
4³ = 64
5³ = 125
6³ = 216
7³ = 343
8³ = 512
9³ = 729 ← this number is over 684 → so we keep the previous number: 512
λ³ + 17λ = 648
λ³ + 17λ = 512 + 136
λ³ - 512 + 17λ - 136 = 0
λ³ - 8³ + 17λ - 136 = 0
(λ³ - 8³) + (17λ - 136) = 0
(λ³ - 8³) + 17.(λ - 8) = 0 → recall: (a³ - b³) = (a - b).(a² + ab + b²)
(λ - 8).(λ² + 8λ + 64) + 17.(λ - 8) = 0
(λ - 8).[(λ² + 8λ + 64) + 17] = 0
(λ - 8).(λ² + 8λ + 81) = 0
First case: (λ - 8) = 0 → λ = 8
Second case: (λ² + 8λ + 81) = 0
Δ = 8² - (4 * 81) = 64 - 324 = - 260 = 260i² = 65 * 4i²
λ = (- 8 ± 2i√65)/2
λ = - 4 ± i√65
We keep only the more convenient to continue → λ = 8
(x² + λ)² - [2λx² + 72x + (λ² + 17)] = 0 → where: λ = 8
(x² + 8)² - [16x² + 72x + (64 + 17)] = 0
(x² + 8)² - (16x² + 72x + 81) = 0
(x² + 8)² - (4x + 9)² = 0 ← 2 squares → a² - b² = (a +b).(a - b)
[(x² + 8) + (4x + 9)].[(x² + 8) - (4x + 9)] = 0
(x² + 8 + 4x + 9).(x² + 8 - 4x - 9) = 0
(x² + 4x + 17).(x² - 4x - 1) = 0
First case: (x² + 4x + 17) = 0
Δ = (4)² - (4 * 17) = 16 - 68 = - 52 = 52i² = 13 * 4i²
x = (- 4 ± 2i√13)/2
→ x = - 2 ± i√13
Second case: (x² - 4x - 1) = 0
Δ = (- 4)² - (4 * - 1) = 16 + 4 = 20 = 4 * 5
x = (4 ± 2√5)/2
→ x = 2 ± √5
(X2+ax+B)*(X2+cx+d)=x4-72x-17
So a=-c, d+ac+B=0, ad+BC=-72, bd=-17 ...
@@germanalexandru Did you solve for a, b, c, and d?
@@brianwade4179 It's a simple task (17,-1, 4,-4)). The start ideea is the problem.
@@germanalexandru I didn't find it to be simple. I started with the 4x4 system you stated, which I checked and agreed with. By substitutions to isolate b I arrived at a sixth-degree (b^6) polynomial in b. Using the Rational Root Theorem I found (b+1) and (b-17) to be factors of it. The remaining fourth-degree (b^4) polynomial did not have rational roots. Owing to symmetry I could choose either -1 or +17 for b, so I chose -1. This gave d=17 and in turn a=-4 and c=4. So {a,b,c,d} = {-4,-1,4,17} is a solution. By symmetry {4,17,-4,-1} is another solution. If you have a way to solve that 4x4 without generating a horrendous polynomial, I would like to see it. Once I knew I had (x^2-4x-1)(x^2+4x+17)=0 the roots came right out: x = 2+-sqrt(5) and x = -2+-sqrt(13)i.
@brianwade4179 IT can only be +17 and -1, because the sum of them must be 4^2 and can not be - 16 or You get "i". Interesting is what if it is a valid choice. If I remember I will test it.
Good morning, dear colleague !
The general strategy is good and natural : to get a perfect square in each member from an unknown trinomial.
On the other hand, factorization by (n - 8) is laborious. As always, too many details...