# 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 $1(x)=x^{2}-3 x-2$, find a quadratic polynomial whose zeros are $\frac{1}{2 \alpha+\beta}$ and $\frac{1}{2 \beta+\alpha}$.

Solution:

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-3 x-2$

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

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

$\alpha+\beta=-\left(\frac{-3}{1}\right)$

$\alpha+\beta=-(-3)$

$\alpha+\beta=3$

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

$\alpha \beta=\frac{-2}{1}$

$\alpha \beta=-2$

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

$S=\frac{1}{2 \alpha+\beta}+\frac{1}{2 \beta+\alpha}$

Taking least common factor then we have ,

$S=\frac{1}{2 \alpha+\beta} \times \frac{2 \beta+\alpha}{2 \beta+\alpha}+\frac{1}{2 \beta+\alpha} \times \frac{2 \alpha+\beta}{2 \alpha+\beta}$

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

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

$S=\frac{3 \beta+3 \alpha}{4 \alpha \beta+2 \beta^{2}+2 \alpha^{2}+\beta \alpha}$

$S=\frac{3(\beta+\alpha)}{5 \alpha \beta+2\left(\alpha^{2}+\beta^{2}\right)}$

$S=\frac{3(\beta+\alpha)}{5 \alpha \beta+2\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]}$

By substituting $\alpha+\beta=3$ and $\alpha \beta=-2$ we get,

$S=\frac{3(3)}{5(-2)+2\left[(3)^{2}-2 \times-2\right]}$

$S=\frac{9}{-10+2(13)}$

$S=\frac{9}{-10+26}$

$S=\frac{9}{16}$

$P=\frac{1}{2 \alpha+\beta} \times \frac{1}{2 \beta+\alpha}$

$P=\frac{1}{(2 \alpha+\beta)(2 \beta+\alpha)}$

$P=\frac{1}{4 \alpha \beta+2 \beta^{2}+2 \alpha^{2}+\beta \alpha}$

$P=\frac{1}{5 \alpha \beta+2\left(\alpha^{2}+\beta^{2}\right)}$

$P=\frac{1}{5 \alpha \beta+2\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]}$

By substituting $\alpha+\beta=3$ and $\alpha \beta=-2$ we get,

$P=\frac{1}{5(-2)+2\left[(3)^{2}-2 \times-2\right]}$

$P=\frac{1}{10+2[9+4]}$

$P=\frac{1}{10+2(13)}$

$P=\frac{1}{-10+26}$

$P=\frac{1}{16}$

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

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

$f(x)=k\left(x^{2}-\frac{9}{16} x+\frac{1}{16}\right)$

Hence, the required equation is $f(x)=k\left(x^{2}-\frac{9}{16} x+\frac{1}{16}\right)$ Where $k$ is any non zero real number.