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 $
$\Rightarrow 2 x^{2}+2 x-1 =0$