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
[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
X+1/x= 4 or x^ 2-4x+1= 0
(X- 2)^ 2=3 or x= +-√3+(2)
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
[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
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