I Solved A Polynomial Equation in Three Ways
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- Опубліковано 8 вер 2024
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x=-1/2 is a solution by inspection. Expanding the parenthesis with the binomial theorem and subtracting x^4 from both sides gets us a cubic polynomial. We divide this by (x+1/2) using polynomial long division to obtain a quadratic polynomial. Using the quadratic formula gets us the remaining two solutions.
Thinking in terms of complex numbers in polar form, if x = re^(i theta), then x^4 = r^4 e^(i * 4 theta). If this equals (x+1)^4, then x, x+1 have the same magnitude. Which means that the real part of x is -1/2.
To wit: if x = a + bi, ||x||^2 = a^2 + b^2, and ||x+1|| = (a+1)^2 + b^2, so a^2 = (a+1)^2, i.e. a = -1/2.
We then need look for values of b such that the the change of theta induced by adding 1 is made irrelevant when theta is multiplied by 4. This suggests theta is some multiple of pi/4, which, if a = -1/2, gives us theta = 3 pi/4 and theta = 5 pi/4. (Along with the obvious theta = pi)
The three roots are -1/2 + i/2, -1/2, and -1/2 - i/2.
This is a problem where I think it's useful to consider the geometry of complex numbers.
I also like the solution that starts by dividing by x^4, which pulls out the 4th roots of unity. Related in spirit.
I dislike algebraic solutions that seem to require foreknowledge of what the target polynomial is.
I've collected many dozens of bookmarks to math channels here on UA-cam. But it turns out that almost the only one I return to daily is this one! I like the thought problems and I learn alot from them. Some day I might perhaps even find use for any of this ;-D
Glad to hear that! 🤩
x^2>=0. function f(x) = x^2 is increasing for x >=0. it means if f(x1)=f(x2) than x1=x2. for any x1, x2>=0.
we can write original equation as f((x+1)^2)= f(x^2). From here you get (x+1)^2=x^2 which leads to linear equation.
You meant 8 is a perfect cube, it’s not a perfect square.
Set y = x + 1/2, x = y - 1/2
Then (y + 1/2)⁴ = (y - 1/2)⁴
One can use the binomial expansion:
4y³/2 + 4y/2³ = -4y³/2 - 4y/2³
y(y² + 1/4) = 0
Nice
This is what I did too
A fourth method, that I used, is a different direction from the second method. (x+1)^4-x^4=((x+1)^2-x^2)((x+1)^+x^2)=(x+1-x)(x+1+x)((x+1)^2-i^2x^2)=1*(2x+1)*(x+1-ix)*(x+1+ix). Three solutions where each parenthesis equals to one :)
We know x=0 is not a solution. So divide both sides by x^4:
(1+1/x)^4 = 1
This gives us four equations:
1+1/x = 1 -> no solutions
1+1/x = i -> x = -1/2 - i/2
1+1/x = -1 -> x = -1/2
1+1/x = -i -> x = -1/2 + i/2
So our three solutions are -1/2, -1/2 + i/2, and -1/2 - i/2. The original equation is a cubic in disguise so we know these are the only solutions
nice. I thought about using this but then changed my mind. Really good way to approach it
Make the substitution t = x + 1/2. This creates a nice symmetry which allows for cancellation after expanding the binomials. I got t (4 t^2 + 1) = 0, which gives the same x:s
Excellent, as always, and even though this was an easy one, I learned something, which you don't do every day, at 80. I used difference of two squares (twice) + quadratic equation formula for the other factor.
if we root both sides , (x+1)^2 = x^2 , from there 2x = -1 , x = -1/2
(x+1)^4=x^4 -> x+1=x or x+1=-x -> x=-1/2
other solution : geometrical. x^4 is an odd function, the only value for which we have x+1 and x having the same image through x^4 is -1/2
First of all: missing assumption concerning whether x is real or complex.
If x is real the solution is simple - just try to imagine yourself 4th order parabola (x+1)⁴ and the second one x⁴.
(x+1)⁴ is shifted left from 0, so minimum is at point (-1,0), the second one is just regular 4th order curve (parabola) with minimum at (0,0)
Now you can imagine the common point, the section between 1st and 2nd curve - it must be between these minima, so -1
I Solved A Polynomial Equation: (x + 1)⁴ = x⁴; x = ?
First method:
(x + 1)⁴ - x⁴ = 0, [(x + 1)² - x²][(x + 1)² + x²] = (2x + 1)(2x² + 2x + 1) = 0
2x + 1 = 0; x = - 1/2 or 2x² + 2x + 1 = 0; x = (- 1 ± i)/2; Complex roots
Second method:
Let: y = x + 1, y - x = 1; (x + 1)⁴ - x⁴ = y⁴ - x⁴ = 0
(y² - x²)(y² + x²) = (y - x)(y + x)(y² + x²) = (1)(y + x)(y² + x²) = (y + x)(y² + x²) = 0
y + x = 0, (x + 1) + x = 0 or y² + x² = 0, (x + 1)² + x² = 0, 2x² + 2x + 1 = 0
x = - 1/2 or x = (- 1 ± i)/2
Answer check:
x = - 1/2, (x + 1)⁴ = (- 1/2 + 1)⁴ = (1/2)⁴ = (- 1/2)⁴ = x⁴; Confirmed
x = (- 1 ± i)/2, 2x² + 2x + 1 = 0, (x + 1)² = - x²; (x + 1)⁴ = (- x²)² = x⁴; Confirmed
Final answer:
x = - 1/2, x = (- 1 ± i)/2 or x = (- 1 - i)/2
My solution to this problem was much simpler.
Start with the problem:
(x + 1) ^ 4 = x ^ 4
Take the fourth root:
x + 1 = x
Subtract x from both sides:
1 = 0
Tah-da!!! Amazing!!!
😂😂😂😂😂😂😂😂
Great! 🤣
I used the difference of 2 squares, it was the most obvious way.....
x = -1/2
have you kept track of number of problems you have solved in your life time?
No I haven't
I used RRT.