[Time slot 3-56 onwards Application of a^2 - b^2=(a+b)(a-b) cancels the denominator and for such a solution sir, u did not get another quadratic equation ] Here is the solution We may get the following equation when we divide the numerator and denominator by x ^7 (x^2 +1/x^2 +1)/(x +1/x +1) =3 > (x +1/x)^2 -2x*1/x +1)/(x +1/x +1) =3 >(a^2 -1)/(a+1)=3 ( **pl note that here we did not use a^2 - 1=(a+1)(a-1) to cancel out the denominator) >a^2-1=3a +3 > a^2 -3a -4= 0 >(a -4)(a+1)=0 Hence a=4 or -1 When a =4 then x + 1/x =4 >x ^2 -4x +1= 0 x = 2 +/- √3 If a = -1 then x +1/x =-1 x = (-1+/- √3i )/2
a=-1 is not a valid solution of the original equation, just like x=0 is not. The left hand side are undefined for exactly these values. The limits as x->0 or a->-1 do not satisfy the equation either.
@@kareolaussen819 May I request u to watch the video from 1-20 to 1-34? We see a reduced form of the given equation in this time slot. If we cross multiply this reduced form, we will get An equation of degree 4 that means we will get 4 roots. You cited logic to refute 4 solutions and advocated for two solutions. I think that u are thinking of my four solutions will be meaningful at a time. But such solution states that at any single time one and only one solution is accepted.
@@PrithwirajSen-nj6qq By cross multiplying you introduce false solutions: (a^2 - 1)/(a+1) = (a-1)(a+1)/(a+1) = 3. The left hand side (LHS) is undefined for a=-1. But lim_{a - >-1} LHS = -2, different from 3.
Divide numerator and denominator by x^7, and introduce u=x+1/x to get equation (u^2-1)/(u+1)=u-1=3. Next solve x+1/x=4 or (x-2)^2=4-1=3 to get x = 2 ± √3
@@ΕυριπίδηςΛόηςSure I can! The left hand side is undefined when u=-1 and x=0, so these do not lead to a valid solutions of the original equation. By rewriting the equation to polynomial form one introduces false roots.
X+1/x= 4 or x^ 2-4x+1= 0
(X- 2)^ 2=3 or x= +-√3+(2)
[Time slot 3-56 onwards
Application of a^2 - b^2=(a+b)(a-b) cancels the denominator and for such a solution sir, u did not get another quadratic equation ]
Here is the solution
We may get the following equation when we divide the numerator and denominator by x ^7
(x^2 +1/x^2 +1)/(x +1/x +1) =3
> (x +1/x)^2 -2x*1/x +1)/(x +1/x +1) =3
>(a^2 -1)/(a+1)=3 ( **pl note that here we did not use a^2 - 1=(a+1)(a-1) to cancel out the denominator)
>a^2-1=3a +3
> a^2 -3a -4= 0
>(a -4)(a+1)=0
Hence
a=4 or -1
When a =4
then x + 1/x =4
>x ^2 -4x +1= 0
x = 2 +/- √3
If a = -1
then x +1/x =-1
x = (-1+/- √3i )/2
a=-1 is not a valid solution of the original equation, just like x=0 is not. The left hand side are undefined for exactly these values. The limits as x->0 or a->-1 do not satisfy the equation either.
@@kareolaussen819
May I request u to watch the video from 1-20 to 1-34?
We see a reduced form of the given equation in this time slot.
If we cross multiply this reduced form, we will get
An equation of degree 4 that means we will get 4 roots.
You cited logic to refute 4 solutions and advocated for two solutions.
I think that u are thinking of my four solutions will be meaningful at a time.
But such solution states that at any single time one and only one solution is accepted.
@@PrithwirajSen-nj6qq By cross multiplying you introduce false solutions:
(a^2 - 1)/(a+1) = (a-1)(a+1)/(a+1) = 3.
The left hand side (LHS) is undefined for a=-1. But lim_{a - >-1} LHS = -2, different from 3.
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Divide numerator and denominator by x^7, and introduce u=x+1/x to get equation
(u^2-1)/(u+1)=u-1=3.
Next solve x+1/x=4 or (x-2)^2=4-1=3 to get
x = 2 ± √3
You should not cancel u+1
@@ΕυριπίδηςΛόηςSure I can! The left hand side is undefined when u=-1 and x=0, so these do not lead to a valid solutions of the original equation. By rewriting the equation to polynomial form one introduces false roots.