Question

Find the quadratic polynomial, the sum of whose zeros is $2$ and their product is $-12.$ Hence, find the zeros of the polynomial.

Answer 1

Let $\alpha$ and $\beta$ be the zeros of the required polynomial $f\left(x\right)$.
Then given $\alpha +\beta =2$ and $\alpha \times \beta =-12$
Thus, required polynomial with zeros $\alpha$ and $\beta$ is
$f\left(x\right)=x^2-(\alpha +\beta )x+(\alpha \times \beta )$
$\Rightarrow f\left(x\right)=x^2-2x+(-12)$
$\Rightarrow f\left(x\right)=x^2-2x-12$
Hence, the required polynomial is $f\left(x\right)=x^2-2x-12$.
To find roots of $f\left(x\right)=x^2-2x-12$:
Use quadratic formula, $x=\frac{-b\pm \sqrt{(b{)}^2-4ac}}{2a}$
Here, $a=1,b=-2,c=-12$
$\Rightarrow x=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(1\left)\right(-12)}}{2\left(1\right)}$
$\Rightarrow x=\frac{2\pm \sqrt{4+48}}{2}$
$\Rightarrow x=\frac{2\pm \sqrt{52}}{2}$
$\Rightarrow x=\frac{2\pm \sqrt{13\times 2^2}}{2}$
$\Rightarrow x=\frac{2\pm 2\sqrt{13}}{2}$
$\Rightarrow x=\frac{2(1\pm \sqrt{13})}{2}$
$\Rightarrow x=1\pm \sqrt{13}$
$\therefore x=1+\sqrt{13}$ and $x=1-\sqrt{13}$