You can use a substitution b = a - 1/2 Then: (b-1/2)^4 = (b+1/2)^4 Developing the equation all even power coefficient cancels out and you get: 4b^3+b=0 b(4b^2+1)=0 Which can be easily solved obtaining b = 0 or i/2 or -i/2 Using a=b+1/2 you obtain the solutions
How about: a^4 = (a-1)^4 => ((a-1)^4)/(a^4) = 1 => ((a-1)/a)^4 = 1 => (a-1)/a = 1, -1, i, -i (4th roots of 2) Case 1: (a-1)/a = 1 - no such a Case 2: (a-1)/a = -1 => a = 1/2 Case 3: (a-1)/a = i => a = (1+i)/2 Case 4: (a-1)/a = -i => a = (1-i)/2
I just used the binomial theorem and evaluated the cubic for real roots and then found the other roots using the quadratic formula, took me 4 minutes without pen and paper needed
@@willemesterhuyse2547 that is interesting, and quite new for me since at school we haven't done a lot with anything to the power of more than 2 and sometimes 3, so it's very unintuitive for me that you can't just do the 4th root to turn a parameter/variable into its base form from a "powered" form
You can use a substitution b = a - 1/2
Then: (b-1/2)^4 = (b+1/2)^4
Developing the equation all even power coefficient cancels out and you get:
4b^3+b=0
b(4b^2+1)=0
Which can be easily solved obtaining b = 0 or i/2 or -i/2
Using a=b+1/2 you obtain the solutions
Excellent!!!
How about:
a^4 = (a-1)^4 => ((a-1)^4)/(a^4) = 1
=> ((a-1)/a)^4 = 1
=> (a-1)/a = 1, -1, i, -i (4th roots of 2)
Case 1: (a-1)/a = 1 - no such a
Case 2: (a-1)/a = -1 => a = 1/2
Case 3: (a-1)/a = i => a = (1+i)/2
Case 4: (a-1)/a = -i => a = (1-i)/2
Excellent delivery 👏
I just used the binomial theorem and evaluated the cubic for real roots and then found the other roots using the quadratic formula, took me 4 minutes without pen and paper needed
a⁴ = (a - 1)⁴
a⁴ - (a - 1)⁴ = 0
[a²]² - [(a - 1)²]² = 0 → recall: a² - b² = (a + b).(a - b)
[a² + (a - 1)²].[a² - (a - 1)²] = 0
[a² + (a² - 2a + 1)].[a² - (a² - 2a + 1)] = 0
[a² + a² - 2a + 1].[a² - a² + 2a - 1] = 0
[2a² - 2a + 1].[2a - 1] = 0
First case: [2a - 1] = 0
2a - 1 = 0
2a = 1
→ a = 1/2
Second case: [2a² - 2a + 1] = 0
2a² - 2a + 1 = 0
Δ = (- 2)² - (4 * 2 * 1) = 4 - 8 = - 4 = 4i²
a = (2 ± 2i)/4
→ a = (1 ± i)/2
Fantabulous! 👏
Why not square root both sides?
If the initial value is equal then their square root must be equal too.
Note that, starting as given we cannot draw the positive 4th root on both sides, because this gives an inconsistent equation.
Wait so you can't just make a = a - 1 ?
Why?
@@Eevneon Because then you can minus a on both sides to get 0 = -1: inconsistent.
@@willemesterhuyse2547 that is interesting, and quite new for me since at school we haven't done a lot with anything to the power of more than 2 and sometimes 3, so it's very unintuitive for me that you can't just do the 4th root to turn a parameter/variable into its base form from a "powered" form
which class is this one
which class is this one