\(x^2-3x+1=0\) হলে \(x^2-\frac{1}{x^2}\) এর মান কত?

  • 5√3
  • 3√5
  • 4√5
  • 6√5

\( x^2-3x+1=0\\ = x^2-3x=-1\\ = x(x-3)=-1\\ = x-3=-\frac{1}{x}\\ = x+\frac{1}{x}=3\\ = (x+\frac{1}{x})^2=3^2\\ = (x+\frac{1}{x})^2=9\\ = (x-\frac{1}{x})^2+4.x.\frac{1}{x}=9\\ = (x-\frac{1}{x})^2=9-4\\ = (x-\frac{1}{x})^2=5\\ = x-\frac{1}{x}=√5 এখন, x^2-(\frac{1}{x})^2=(x+\frac{1}{x}) (x-\frac{1}{x})\\ =3√5\)