6(x2+1/x2)-35(x-1/x)+62 solve for x Share with your friends Share 0 Abhishek Jha answered this Dear Student, Here is the solution of your asked query: 6x2+1x2-35x-1x+62=0Taking x-1x=y ...(i) , we have;x-1x2=y2⇒x2+1x2-2×x×1x=y2⇒x2+1x2-2=y2⇒x2+1x2=y2+2 ...(ii)Substituting (i) and (ii) in the given equation we get;6y2+2 -35y+62=0⇒6y2+12-35y+62=0⇒6y2-35y+74=0⇒y=--35±-352-4×6×742×6 =35±55112=35±2312 {Since 551=23.4733, taking approximate value around 23}=35+2312 and 35-2312=5812 and 1So we have;x-1x=1 and x-1x=5812⇒x2-1=x and x2-1=5812x⇒x2-x-1=0 and 12x2-12=58x⇒x2-x-1=0 and 12x2-58x-12=0⇒x=--1±-12-4×1×-12×1 and x=--58±-582-4×12×-122×12⇒x=1±52 and x=58±394024 Regards 0 View Full Answer