Question

jisme hai dum solve it...vahi h maths ka asli singham...

Answer 1

2(x^2 +1/x^2) +3( x - 1/x) - 4= 0

2( (x - 1/x)^2 + 2) +3 ( x - 1/x) - 4= 0

2(x - 1/x)^2 + 3(x - 1/x)= 0

(x - 1/x)(2x - 2/x +3)= 0

so

x - 1/x = 0

x = 1 and -1

and

2x - 2/x + 3 = 0

2x^2 + 3x -2 = 0

2x^2 +4x - x - 2=0

2x(x +2)-1(x+2)=0

(x+2)(2x -1)=0

x= -2 and +1/2

so the value of x is 1 ,-1,-2,1/2

using

(a + b)^2 = a^2 + 2ab + b^2