easy to see that the equation is symmetrical to x=2. let's use substitution t = x-2 . so x=t+2 (t+2)^4 = (t-2)^2 t^4+8t^3+24t^2+32t+16 = t^4−8t^3+24t^2−32t+16 8t^3+32t = -8t^3−32t 16*t^3+48*t=0 t*(t^2+4)=0 t= 0, t = 2i, t =-2i x=2, x= 2+2i, x=2-2i
The solutions come from the 4 equations "X=(X-4) times A" where A are the 4th roots of 1, i.e. members of the set {1, i, -1, -i}. For A=1 the equation X=X-4 has no solution. The other equations are X=i(X-4), X=(4-X) , AND X=i(4-X). The solutions of these equations are X=2-2i, X=2, and X=2+2i .
Rule: Sqrt (a^4) = +/- a^2. So: sqrt (x^4)=+/-sqrt ((x-4)^4) =>x^2=+/- (x-4)^2 With plus-sign: Quadratic, x=2 With minus-sign: x^2=-(x-4)^2=-(x^2+4^2-8x)=-x^2+16x-16 =>2x^2-16x+16=0 or so Solve for x and get the complex roots as stated.
Not a more efficient way, but interesting to me: Observe that we are comparing even powers of 2 numbers, one smaller than the other. In this case, that means that x is 1/2 of 4, so that x-4 = -2 and (-2)^4 = 2^4. Therefore x=2 is one solution. Next, bring (x-4)^4 to the lhs and expand the polynomial x^4 - (x-4)^4. Do the subtraction and observe that the 4th power terms cancel, leaving us with a 3rd order polynomial. There are 3 roots and one of them is x=2. What are the remaining factors of the 3rd order poly? Divide it by the (x-2) factor, which leaves an exact quotient, x^2 -4x + 8. Now find its 2 roots with the quadratic formula.
It looks like a standard exercise from a textbook or a lecture. x^4=(x-4)^4 x^4/(x-4)^4=1 => x=4*e^((i*2*k*pi)/4)/(e^((i*2*k*pi)/4) -1), with k=1,2,3 only. k=/0. x=2, x=2-2i, x=2+2i.
question: could be acceptable to consider x=infinite, as the 4th root?. In fact x=0 and x-4=0 are two parallel lines, which have a common point in infinite.
This solution is unnecessary if we only care about real solutions. When you take 4th root you need to remember to add the ± symbol. We look at the + case: x = x - 4. This is an extraneous root and we reject. Now the - case: x = -x + 4. => 2x = 4, x = 2. QED.
When you open the binomial expansion of (x-4)⁴ you get a term having power 4. This term will be cancelled by the x⁴ on the left side of the equality. So that's why this equation have 3 roots (including 1 real and 2 complex ).
A quartic equation with only 3 roots... 🤔 It wasn't really a quartic equation, it was a cubic equation in disguise. Like taking a linear equation and adding a x^2 to both sides doesn't make it a quadratic. It is still linear.
After expanding the right side of the equation, then subtract the fourth-order term from each side, we have a third-degree equation with (at most) three solutions.
easy to see that the equation is symmetrical to x=2. let's use substitution t = x-2 . so x=t+2
(t+2)^4 = (t-2)^2
t^4+8t^3+24t^2+32t+16 = t^4−8t^3+24t^2−32t+16
8t^3+32t = -8t^3−32t
16*t^3+48*t=0
t*(t^2+4)=0
t= 0, t = 2i, t =-2i
x=2, x= 2+2i, x=2-2i
(t+2)⁴-(t-2)⁴=( (t+2)²+(t-2)² )( (t+2)²-(t-2)² ) = ( 2t² + 8 )( 8t )
same symmetrical method for x(x+2)(x+4)(x+6)=9
x⁴ = (x-4)⁴
(x-0)⁴ = (x-4)⁴
[(x-4+4)²]² = [(x-4)²]²
Posons x-4= X
on a[ (X+4)²] ² = [(X)²]²
|(X+4)²]| = |[(X)²]|
(X+4)² = X²
8X +16=0
X = -2
x-4 =-2
x=2
The solutions come from the 4 equations "X=(X-4) times A" where A are the 4th roots of 1, i.e. members of the set {1, i, -1, -i}. For A=1 the equation X=X-4 has no solution. The other equations are X=i(X-4), X=(4-X) , AND X=i(4-X). The solutions of these equations are X=2-2i, X=2, and X=2+2i .
Rule: Sqrt (a^4) = +/- a^2.
So: sqrt (x^4)=+/-sqrt ((x-4)^4)
=>x^2=+/- (x-4)^2
With plus-sign: Quadratic, x=2
With minus-sign:
x^2=-(x-4)^2=-(x^2+4^2-8x)=-x^2+16x-16
=>2x^2-16x+16=0 or so
Solve for x and get the complex roots as stated.
Not a more efficient way, but interesting to me: Observe that we are comparing even powers of 2 numbers, one smaller than the other. In this case, that means that x is 1/2 of 4, so that x-4 = -2 and (-2)^4 = 2^4. Therefore x=2 is one solution. Next, bring (x-4)^4 to the lhs and expand the polynomial x^4 - (x-4)^4. Do the subtraction and observe that the 4th power terms cancel, leaving us with a 3rd order polynomial. There are 3 roots and one of them is x=2. What are the remaining factors of the 3rd order poly? Divide it by the (x-2) factor, which leaves an exact quotient, x^2 -4x + 8. Now find its 2 roots with the quadratic formula.
Consider this:
x^2=+-(x-4)^2
Case1: x^2=(x-4)^2
x=+-(x-4)
x=+(x-4)=>0=-4=>rejected
x=-(x-4) =>x=2
Case2:x^2=-(x-4)^2
x=+-i(x-4)
x=-4/(1+-i)
x=-2+-2i
(x-4)^2 = +- x^2
(x-4) = +- x , +- xi
x = 2, 4/(1-i), 4/(1+i)
x = 2, 2+-2i
@@cosmolbfu67 love it!
Quickest way by far is to take the square root twice. And don't forget that there are 2 square roots!
It looks like a standard exercise from a textbook or a lecture. x^4=(x-4)^4 x^4/(x-4)^4=1 => x=4*e^((i*2*k*pi)/4)/(e^((i*2*k*pi)/4) -1), with k=1,2,3 only. k=/0. x=2, x=2-2i, x=2+2i.
-> by faktoring , x^3-6x^2+16x-16=0 , -> (x-2)(x^2-4x+8)=0 , x= 2 , 2+2i , 2-2i , test , 2^4=16 , (x-4)^4=(-2)^4 , (-2)^4=16 , same , OK ,
to took the 4 th. root or high 1/4 is o.k. with the statement : "and x and (x-4) are positive".
but then you change the set of kandidates.
128x^4 4^4x^4 2^2^2^2^x^2^2 1^1^1^1^x^1^2 x1^2 (x ➖ 2x+1).
x^2-(x-4)^2= (x-x+4)(x+x-4)=8(x-2), as you have just used this formula. No need to calculate (x-4)^2.
question: could be acceptable to consider x=infinite, as the 4th root?. In fact x=0 and x-4=0 are two parallel lines, which have a common point in infinite.
This solution is unnecessary if we only care about real solutions. When you take 4th root you need to remember to add the ± symbol. We look at the + case: x = x - 4. This is an extraneous root and we reject. Now the - case: x = -x + 4. => 2x = 4, x = 2. QED.
# "if we only care about real solutions"
...then we are back in the 16th Century.
@@moebadderman227 no because the question is specifically asking for x∈ℝ. It’s a math problem not a problem for real life engineering.
imaginary number is also REAL. literally, in many of the technologies you enjoy in your life.
@@y0ng_y0utube
# "imaginary number is also REAL"
wat
Same thing there should be 4 roots right ?explain me
When you open the binomial expansion of (x-4)⁴ you get a term having power 4.
This term will be cancelled by the x⁴ on the left side of the equality.
So that's why this equation have 3 roots (including 1 real and 2 complex ).
@@pinkipensia4671 😯 thanks
@@pinkipensia4671 Thanks. Would have been nice if he'd explained this.
The fourth root is the solution of x=x-4
@@lunstee Good guess. Wrong answer.
x^2=(x-4)^2
x^2=x^2-8x+16
8x=16
x=2
Answer is simply X=2 , btw You missed the negative root when you got 4th root of equation in first part or you wouyhave go lt x=2
4-x=±x{i;1}
x=4/{2; 1±i}
x=2{1 ; 1±i}
x=y+2;
y^4+4*y^3*2+6*y^2*2^2+4*y*2^3+2^4 = y^4-4*y^3*2+6*y^2*2^2-4*y*2^3+2^4
y^3*2+y*2^3 =0;
y*(y^2+2^2)=0
=> y =0, 2i, -2i => x =2, 2+2i, 2-2i
A quartic equation with only 3 roots... 🤔
It wasn't really a quartic equation, it was a cubic equation in disguise.
Like taking a linear equation and adding a x^2 to both sides doesn't make it a quadratic. It is still linear.
We are supposed to have 4 roots. What is the forth root?
The true solution is in réel number .
9:12 that fly 😂😂
Is it 2 ?
I don't understand why there's no 4 solutions, why just 3.
After expanding the right side of the equation, then subtract the fourth-order term from each side, we have a third-degree equation with (at most) three solutions.
Is this an actual Olympiad question, because I solved it in 5 minutes😅
I mean it is still a hard question but shouldn't it be a bit harder?
Can a polynomial of degree n have n+1 solutions? Is this possible in mathematics ? Really?
Nice
x⁴=(x-4)⁴
(x-4)⁴-x⁴=0
[(x-4)²]²-(x²)²=0
[(x-4)²+x²][(x-4)²-x²]=0
(x-4)²+x²=0
x²-8x+16+x²=0
2x²-8x+16=0
x²-4x+8=0
x²-4x+4=-4
(x-2)²=-4
|x-2|=2i
x-2=±2i
x=2±2i ❤❤
(x-4)²-x²=0
x²-8x+16-x²=0
16-8x=0
-8x=-16
x=2 ❤
2.
2
First for all n > 3
3th FIRST!!!
First third
First second
First!!!!!
First 🥇🥇🥇🥇🥇