<span>Which of the following is the solution set of the equation 10(x^2+1/x^20)-63(x-1/x)=52=0 I think you type wrongly the equation: It is : </span><span>10(x^2+1/x^2)-63(x-1/x)+52=0 We know: x</span>²+1/x²=(x-1/x)²+2, pose x-1/x=t and we have x²+1/x²=t²+2 So 10(t²+2)-63t+52=0 or 10t²-63t+72=0 we have t=(63+33):20= 24/5 and t=(63-33):20=3/2
+ If t=3/2 we have x-1/x=3/2 or 2x²-3x-2=0 and we have x=2, x=-1/2 + Of t=24/5 we have x-1/x=24/5 or 5x²-24x-5=0 or x=5 and x=-1/5
