Question

Form the quadratic equation whose roots are $(-3+2\sqrt{5}),(-3-2\sqrt{5})$

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 $m=(-3+2\sqrt{5})$ and $n=(-3-2\sqrt{5})$, therefore,

The sum of the roots is:

$m+n=(-3+2\sqrt{5})+(-3-2\sqrt{5})=-3+2\sqrt{5}-3-2\sqrt{5}=-3-3=-6$

And the product of the roots is:

$mn=(-3+2\sqrt{5})\times (-3-2\sqrt{5})=(-3{)}^2-(2\sqrt{5}{)}^2=9-20=-11(\because a^2-b^2=(a-b\left)\right(a+b\left)\right)$

Therefore, the required quadratic equation is

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

$\Rightarrow x^2-(-6)x+(-11)=0\Rightarrow x^2+6x-11=0$

Hence, $x^2+6x-11=0$ is the quadratic equation whose roots are $(-3+2\sqrt{5})$ and $(-3-2\sqrt{5})$.