Question

$a=-2+\sqrt{3}$ and $b=-2-\sqrt{3}$ then $\frac{1}{a^2}-\frac{1}{b^2}$ is equal to

• A. 14
• B. -14
• C. 8$\sqrt{3}$
  
• D. 7$\sqrt{3}$

Answer 1

The correct option is C 8$\sqrt{3}$

$a=-2\sqrt{3},b=-2-\sqrt{3}$

$b^2-a^2=(b+a)(b-a)=(-2-\sqrt{3}-2+\sqrt{3})(-2-\sqrt{3}+2\sqrt{3}).$

$b^2-a^2=(-4)(-2\sqrt{3})=8\sqrt{3}$

${\left(ab\right)}^2=a^2{b}^2={\left\{\right(-2+\sqrt{3}\left)\right(-2-\sqrt{3}\left)\right\}}^2=1^2=1$

So, $\frac{1}{a^2}-\frac{1}{b^2}=\frac{b^2-a^2}{b^2{a}^2}=\frac{8\sqrt{3}}{1}=8\sqrt{3}$