Question

Find the quadratic equation whose one root is $2+\sqrt{5}$.

• A. $x^2-4x+1=0$
• B. $x^2-4x-1=0$
• C. $x^2-4x+3=0$
• D. $x^2-4x-3=0$

Answer 1

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

One root of the equation is $2+\sqrt{5}$. So, the next root will be $2-\sqrt{5}$

$\therefore$ $x=2+\sqrt{5}$ and $x=2-\sqrt{5}$

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

$\therefore (x^2+(4-5)-2x-\sqrt{5}x-2x+\sqrt{5}x)=0$

$\therefore x^2-4x-1=0$ is the required quadratic equation.