Question

The quadratic equation whose roots are $\frac{1}{3+\sqrt{2}}$ and $\frac{1}{3-\sqrt{2}}$, will be

• A. $7 x^{2}-6 x+1=0$
• B. $6 x^{2}-7 x+1=0$
• C. $x^{2}-6 x+7=0$
• D. $x^{2}-7 x+6=0$

Answer 1

Answer: (A)

Given roots are $\frac{1}{3+\sqrt{2}}$ and $\frac{1}{3-\sqrt{2}}$
i.e. $\frac{1}{(3+\sqrt{2})} \times \frac{(3-\sqrt{2})}{(3-\sqrt{2})}$
and $\frac{1}{(3-\sqrt{2})} \times \frac{(3+\sqrt{2})}{(3+\sqrt{2})}$
$\Rightarrow \frac{3-\sqrt{2}}{7}$ and $\frac{3+\sqrt{2}}{7}$
So, the required quadratic equation is $x^{2}-$ (Sum of the roots )
$x+$ Product of the roots $=0$
$\Rightarrow x^{2}-\left[\frac{3-\sqrt{2}}{7}+\frac{3+\sqrt{2}}{7}\right] x+\left(\frac{3-\sqrt{2}}{7}\right)\left(\frac{3+\sqrt{2}}{7}\right)=0$
$\Rightarrow x^{2}-\left(\frac{6}{7}\right) x+\frac{9-2}{49}=0$
$\Rightarrow x^{2}-\frac{6}{7} x+\frac{7}{49}=0$
$\Rightarrow 7 x^{2}-6 x+1=0$