Question

If one root of a quadratic equation is $\frac{1}{1+\sqrt{3}},$ then the quadratic equation is

• A. $2x^2+x-1=0$
• B. $2x^2-2x-1=0$
• C. $2x^2+2x+1=0$
• D. $2x^2+x+1=0$
• E. $2x^2+2x-1=0$

Answer 1

Answer: (E)

If one root of the quadratic equation is $\frac{1}{1+\sqrt{3}}$,
hen other root will be $\frac{1}{1-\sqrt{3}}$.
Let $\alpha =\frac{1}{1+\sqrt{3}}$ and $\beta =\frac{1}{1-\sqrt{3}}$
Now, $\alpha +\beta =\frac{1}{1+\sqrt{3}}+\frac{1}{1-\sqrt{3}}$
$=\frac{1-\sqrt{3}+1+\sqrt{3}}{1-3}=-1$
Also, $\alpha \beta =\frac{1}{1+\sqrt{3}}\times \frac{1}{1-\sqrt{3}}$
$=\frac{1}{1-3}=-\frac{1}{2}$
Hence, required quadratic equation will be
$x^2+x-\frac{1}{2}=0$
$\Rightarrow 2x^2+2x-1=0$