substitute: y=x-2; a complex y to the power of 4 shall be 16; r=16^(1/4)=2; theta= 0 or π/2 or -π/2 or -π; all 4 thetas lead to 0 if multiplied by 4 (rotated 4 times by theta); to get x shift 4 ys in the complex plane
I' vê just red some comments, agreed. Clear at very first sight on equation. Zero and minus 4. Two solutions is right for this particular equation.....
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
@ Do you think you are teaching me something, I don’t care how many roots there are, there are only two that are of any interest, and I found them by inspection, enough said! If I needed the other roots. And I cannot imagine any reason for needing them, I could determine them by inspection too, but why bother!
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
@ arafaahil8068 No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
Too complicated solution ! 16 is equal to 2^4, then we have the difference of two square, and using the remarquable identities, we get two real solutions and two complex solutions. We have to remark that "x equal 0" is a solution.
There should be four solutions to this. Only two will be real. And the narrator made this way too complex. You could have taken roots on both sides and wound up with (x+2)^2=+/- 4. Take a second root to get x+2= +/- (+/- 4)^1/2. The solutions are 0 and -4. These are your real roots. The imaginary roots will come in when you do the second root of the negative roots of the first square root. The second negative sign will give you your complex conjugate.
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
@@schcl you could get complex directly ... apply sqroot and then (x+2)^2= +/-4, an then you have x^2+4x+8=0 and x^2+4x=0, this Indian boy spent 4 min to get to that
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
substitute: y=x-2; a complex y to the power of 4 shall be 16; r=16^(1/4)=2; theta= 0 or π/2 or -π/2 or -π; all 4 thetas lead to 0 if multiplied by 4 (rotated 4 times by theta); to get x shift 4 ys in the complex plane
I' vê just red some comments, agreed. Clear at very first sight on equation. Zero and minus 4. Two solutions is right for this particular equation.....
x=2×cos(n×pi/2)-2+2i×sin(n×pi/2), n=1, 2, 3, 4
So roots are -4, 0, -2±2i
[(X+2)^4]^1/2 = (16)^1/2
(X+2)^2 = = 4
[(X+2)^2]^1/2 = 4^1/2
X+2 = 2
X = 2-2
X=0
0 and -4
By inspection 2^4 = 16, therefore x = 0, any other roots aren’t worth the time considering, although x = - 4 works too.
solutiom.
((x+2)/2)^4=1
note 1=e^2πni , n=0,1,2,3
((x+2)/2)=1^(1/4) =1,i,-1,-i
x+2=2(1,-1,i,-i)
x=-2+-2=0,-4
x=-2+-2i
4 solutions
x= 0, -4, -2-2i, -2+2i
@
Thank you, but I wasn’t asking for help with a solution which I could have done myself! My point was the total stupidity of the question.
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
@
Do you think you are teaching me something, I don’t care how many roots there are, there are only two that are of any interest, and I found them by inspection, enough said! If I needed the other roots. And I cannot imagine any reason for needing them, I could determine them by inspection too, but why bother!
Dont need complex when there are simple.
〈×+2)⁴=16=2⁴
(x+2)⁴=2⁴
x+2=2......x=0
Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
(x+2)^2=+_4
x+2=+_2. x=0. x=--4
x+2=+_2 i
x=--2+2 i. x= --2 --2i
At 71 took me 5 seconds to solve, x is 0
Absolute value 0; -4
Same here
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
I saw the equation and thought : x=0 and x=-4 .
X=,0
Х1=0 ; Х2=-4
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
16=2power4 and x+2) power 4
comparing x+2 and 2
x+2 = 2 or x = 0
very simple
why all that exhaustion?
@ arafaahil8068 No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
Too complicated solution !
16 is equal to 2^4, then we have the difference of two square, and using the remarquable identities, we get two real solutions and two complex solutions. We have to remark that "x equal 0" is a solution.
There should be four solutions to this. Only two will be real. And the narrator made this way too complex. You could have taken roots on both sides and wound up with (x+2)^2=+/- 4. Take a second root to get x+2= +/- (+/- 4)^1/2. The solutions are 0 and -4. These are your real roots. The imaginary roots will come in when you do the second root of the negative roots of the first square root. The second negative sign will give you your complex conjugate.
X is 0
Absolute value 0; -4
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
X=0
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
Y porqué no 2^4???
You could have literally done this in 3 seconds. 16 is (+/-2)^4, meaning (x+2)=+/-2
It's pretty simple
You are missing the complex solutions
@@schcl you could get complex directly ... apply sqroot and then (x+2)^2= +/-4, an then you have x^2+4x+8=0 and x^2+4x=0, this Indian boy spent 4 min to get to that
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.
X=0
No. It is a 4th degree polynomial equation. So, x has four values. Let x+2 = y ; so, y^4-16=0 ; i,e (y^2+4)(y^2-4)= 0 ; i,e (y+2i)(y-2i)(y+2)(y-2)=0 ; so y= 2i or -2i or 2 or -2 ; i,e x=2i-2 or -2i-2 or 0 or -4.