This equation has only _two_ solutions, not four. The zeros of x² + x + 1 are not solutions of the equation because x² + x + 1 is a factor of the denominator x⁴ + x + 1 so the fraction is undefined whenever x² + x + 1 is zero.
I agree with you. Even with the t substitution the denominator becomes t^2-1 since this equation still represent the original equation before cross multiplication t cannot be equal to +/-1. Thus only 2 solutions.
The other solution is to multiply the denominator and numerator on the left side by (x^2-1). Then, the denominator changes (x^2+x+1) and the numerator changes (x^2-x+1). Finally, this equation can be transformed into 2x^2-5x+2=0 -> x = 2, 1/2.
In the math One question you can solve in many ways but valuable only one which way is simple and small see my solution Suppose this 4 degree equation have rational or integer solution Then (x²+x+1)²/(x⁴+x²+1)=7/3 Now (x²+x+1)²/(x⁴+x²+1)=7²/21 Compare both sides (x²+x+1)²=7² ,(x⁴+x²+1)=21 x²+x+1=±7 , x⁴+x²+1=21 x=2,-3,( -1±√33)/2 ,x=2,-2,±√5i There is 2 is common in both equation Then X=2
By inspection I found that x = 2 is a root of this equation. Then the remaining cubic equation must have a second real root. I would inspect or plot the cubic to find the second real root. Following this I would solve the remaining quadratic equation for the remaining two real or complex roots. This is the most reliable method for me because I don't memorize special tricks.
Let f(x) = (x^2 +x + 1)^2 / (x^4 + x^2 + 1) then f(2) = 7/3 and f(1/2) = 7/3 The complex numbers n1 = (-1- i sqrt(3) )/2 and n2 = (- 1 + i sqrt(3) )/2 are not roots, since f(n1) = 0 and f(n2) =0
This equation has only _two_ solutions, not four. The zeros of x² + x + 1 are not solutions of the equation because x² + x + 1 is a factor of the denominator x⁴ + x + 1 so the fraction is undefined whenever x² + x + 1 is zero.
Exactly. If t is a root of x^2 +x +1 then the original rational expression calculated at t is 0/0 which is an indeterminate form.
I agree that there are only two solutions. I did not understand why x^2 + x + 1 = 0 was considered. Perhaps someone can explain this.
Agreed. The LHS cannot both be poorly defined and equate to a constant.
@@82rahsometimes people make mistakes. Nothing to explain here those complex roots are not roots…
I agree with you.
Even with the t substitution the denominator becomes t^2-1 since this equation still represent the original equation before cross multiplication t cannot be equal to +/-1. Thus only 2 solutions.
Got ‘em all!
The other solution is to multiply the denominator and numerator on the left side by (x^2-1). Then, the denominator changes (x^2+x+1) and the numerator changes (x^2-x+1). Finally, this equation can be transformed into 2x^2-5x+2=0 -> x = 2, 1/2.
Genial
In the math One question you can solve in many ways but valuable only one which way is simple and small see my solution
Suppose this 4 degree equation have rational or integer solution
Then
(x²+x+1)²/(x⁴+x²+1)=7/3
Now
(x²+x+1)²/(x⁴+x²+1)=7²/21
Compare both sides
(x²+x+1)²=7² ,(x⁴+x²+1)=21
x²+x+1=±7 , x⁴+x²+1=21
x=2,-3,( -1±√33)/2 ,x=2,-2,±√5i
There is 2 is common in both equation
Then
X=2
BUT YOU LOST THE 2nd SOLUTION x = 0.5
x={1/2,2}
x to the power 2 is the same x square?
Öyle değil mi?
X=2
x = 2
x^4+2x^3+3x^2+2x+1=7n
x^4+x^2+1=3n
By inspection I found that x = 2 is a root of this equation. Then the remaining cubic equation must have a second real root. I would inspect or plot the cubic to find the second real root. Following this I would solve the remaining quadratic equation for the remaining two real or complex roots. This is the most reliable method for me because I don't memorize special tricks.
try harder😊
Think again. The equation has only two solutions, not four as SyberMath claims in the video.
2 real
@@SyberMath no 0/0 cannot be evaluated even for complex numbers so those are not roots
Let f(x) = (x^2 +x + 1)^2 / (x^4 + x^2 + 1) then f(2) = 7/3 and f(1/2) = 7/3 The complex numbers n1 = (-1- i sqrt(3) )/2 and n2 = (- 1 + i sqrt(3) )/2 are not roots, since
f(n1) = 0 and f(n2) =0
X=2, x=0.5, x=(-1±3½i)/2
The last 2'are not accettable
@@giuseppemalaguti435 right😅😅but I like complex numbers..so😅