# If α and β are the zeros of the quadratic polynomial f(x)

Question:

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $l(x)=x^{2}-1$, find the quadratic polynomial whose zeros are $\frac{2 \alpha}{\beta}$ and $\frac{2 \beta}{\alpha}$.

Solution:

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-1$

The roots are $\alpha$ and $\beta$

$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$\alpha+\beta=\frac{0}{1}$

$\alpha+\beta=0$

$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$\alpha \beta=\frac{-1}{1}$

$\alpha \beta=-1$

Let S and P denote respectively the sum and product of zeros of the required polynomial. Then,

$S=\frac{2 \alpha}{\beta}+\frac{2 \beta}{\alpha}$

Taking least common factor we get,

$S=\frac{2 \alpha^{2}+2 \beta^{2}}{\alpha \beta}$

$S=\frac{2\left(\alpha^{2}+\beta^{2}\right)}{\alpha \beta}$

$S=\frac{2[(\alpha+\beta)-2 \alpha \beta]}{\alpha \beta}$

$S=\frac{2[0-2 \times-1]}{-1}$

$S=\frac{2[-2 \times-1]}{-1}$

$S=\frac{2 \times 2}{-1}$

$S=\frac{4}{-1}$

$S=-4$

$P=\frac{2 \alpha}{\beta} \times \frac{2 \beta}{\alpha}$

$P=4$

Hence, the required polynomial $f(x)$ is given by,

$f(x)=k\left(x^{2}-S x+p\right)$

$f(x)=k\left(x^{2}-(-4) x+4\right)$

$f(x)=k\left(x^{2}+4 x+4\right)$

Hence, required equation is $f(x)=k\left(x^{2}+4 x+4\right)$ Where $k$ is any non zero real number.