Question

If $(x+\frac{1}{x})=\frac{10}{3}$, then ${(x-\frac{1}{x})}^2$ is

• A. ${\left(\frac{7}{3}\right)}^2$
• B. ${\left(\frac{8}{3}\right)}^2$
• C. ${\left(\frac{10}{3}\right)}^2$
• D. ${\left(\frac{5}{3}\right)}^2$

Answer 1

The correct option is B ${\left(\frac{8}{3}\right)}^2$

$x+\frac{1}{x}=\frac{10}{3}$

S.O.B.S

${(x+\frac{1}{x})}^2={\left(\frac{10}{3}\right)}^2$

$x^2+\frac{1}{x^2}+2=\frac{100}{9}$

$x^2+\frac{1}{x^2}=\frac{100}{9}-2$

$x^2+\frac{1}{x^2}=\frac{82}{9}$

Let $x-\frac{1}{x}=y$

S.O.B.S

${(x+\frac{1}{x})}^2=y^2$

$x^2+\frac{1}{x^2}-2=y^2$

$\frac{82}{9}-2=y^2$

$\frac{82-18}{9}=y^2$

$\frac{64}{9}=y^2$

i.e; ${(x-\frac{1}{x})}^2=\frac{64}{9}={\left(\frac{8}{3}\right)}^2$