Question

Form the quadratic equation whose roots are $(2+\sqrt{3}),(2-\sqrt{3})$

Answer 1

We know that if $m$ and $n$ are the roots of a quadratic equation $a{x}^2+bx+c=0$, then the sum of the roots is $(m+n)$ and the product of the roots is $\left(mn\right)$. And then the quadratic equation becomes $x^2-(m+n)x+mn=0$

Here, it is given that the roots of the quadratic equation are $m=(2+\sqrt{3})$ and $n=(2-\sqrt{3})$, therefore,

The sum of the roots is:

$m+n=2+\sqrt{3}+2-\sqrt{3}=2+2=4$

And the product of the roots is:

$mn=(2+\sqrt{3})\times (2-\sqrt{3})=2^2-(\sqrt{3}{)}^2=4-3=1(\because a^2-b^2=(a-b)(a+b))$

Therefore, the required quadratic equation is

$x^2-(m+n)x+mn=0$

$\Rightarrow x^2-4x+1=0$

Hence, $x^2-4x+1=0$ is the quadratic equation whose roots are $(2+\sqrt{3})$ and $(2-\sqrt{3})$.