Question

Find the quadratic polynomial, sum of whose zeros is $\left(\frac{5}{2}\right)$ and their product is $1$. Hence, find the zeros of the polynomial?

Answer 1

Let $\alpha ,\beta$ be the zeros of required quadratic polynomial $f\left(x\right)$
We have,
$\alpha +\beta =\frac{5}{2}$,
$\alpha \times \beta =1$

Now, the required polynomial with zeros $\alpha$ and $\beta$ is $f\left(x\right)=x^2-(\alpha +\beta )x+\alpha \times \beta =0$
$\Rightarrow x^2-\left(\frac{5}{2}\right)x+1=0$
$\Rightarrow \frac{x^2-\left(5\right)x+2}{2}=0$
$\Rightarrow 2x^2-5x+2=0$
Splitting the middle term $-5x$ into two terms $-4x$ and $-1x$ such that its product $-4x\times (-1x)=4x^2$ is equals to the product of extreme terms.($2\times 2x^2=4x^2$)
$\Rightarrow 2x^2-4x-1x+2=0$
$\Rightarrow 2x(x-2)-1(x-2)=0$
$\Rightarrow (2x-1)(x-2)=0$
$\Rightarrow 2x-1=0$ and $x-2=0$
$\therefore x=\frac{1}{2}$ and $x=2$
Hence, the zeroes of the polynomial are $\frac{1}{2}$ and $2$.