Question

If α and β are the zeros of the quadratic polynomial f(x) = x² − 3x − 2, find a quadratic polynomial whose zeroes are ${\displaystyle \frac{1}{2\alpha +\beta }+\frac{1}{2\beta +\alpha }}$

Answer 1

Since α and β are the zeros of the quadratic polynomial f(x) = x² − 3x − 2

The roots are α and β

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

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

α + β = -(-3)

α + β = 3

${\displaystyle \alpha \beta =\frac{\text{constant term}}{\text{coefficient of}{x}^2}}$

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

αβ = -2

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

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

Taking least common factor then we have ,

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

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

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

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

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

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

By substituting α + β = 3 and αβ = -2 we get,

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

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

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

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

 

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

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

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

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

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

By substituting α + β = 3 and αβ = -2 we get,

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

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

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

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

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

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

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

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

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