Question

One of the quadratic equation with rational co-ordinates one of whose root is $\frac{1}{2+\sqrt{5}}$, is

• A. $x^2-4x-1=0$
• B. $x^2+4x-1=0$
• C. $x^{2}4x+1=0$
• D. $x^2+4x+1=0$

Answer 1

The correct option is B $x^2+4x-1=0$

observe $\frac{1}{2+\sqrt{5}}=\frac{1}{\sqrt{5}+2}=\left(\frac{1}{\sqrt{5}+2}\right)\left(\frac{\sqrt{5}-2}{\sqrt{5}-2}\right)=\frac{\sqrt{5}-2}{5-4}$

$=\sqrt{5}-2$

$=-2+\sqrt{5}$

consider

Also, the quadratic equation

$(x-(-2+\sqrt{5}\left)\right)(x-(-2-\sqrt{5}\left)\right)=0$

Clearly, $-2+\sqrt{5}$ is a root of two equation and

$(x-(-2+\sqrt{5}\left)\right)(x-(-2-\sqrt{5}\left)\right)$

$=(x+2-\sqrt{5})(x+2+\sqrt{5})$

$=(x+2{)}^2-5$

$=x^2+4+4x-5$

$=x^2+4x-1$

Thus, correct answer is $\left(B\right)$