Question

One of the value of x in the equation 8(x*x+1/x*x)-42(x-1/x)+29 = 0 is

Answer 1

8(x²+1/x²)-42(x-1/x)+29=0

x-1/x=t

वर्ग करने पर

(x-1/x)²=t²

x²+1/x²-2x1/x= t²

t²+2=x²+1/x²

8(t²+2)-42t+29=0

8t²+16-42t+29=0

8t²-42t+45=0

8t²-30t-12t+45=0

2t(4t-15)-3(4t-15)=0

(2t-3)(4t-15)=0

2t-3=0,4t-15=0

t=3/2,15/4

t=x-1/x

3/2= (x²-1)/x

3x=2x²-2

2x²-3x-2=0

2x²-4x+x-2=0

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

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

x= -1/2,2

second t=15/4

15/4=(x²-1)/x

15x=4x²-4

4x²-15x-4=0

4x²-16x+x-4=0

4x(x-4)+1(x-4)=0

(x-4)(4x+1)=0

x=4,-1/4

I am sure this is your answer