Question

If α and are zeroes of a polynomial f(x) = x2- 3x -2, find a quadratic polynomial whose zeroes are 1/2α+β and 1/2β+α.

Answer 1

The required quadratic polynomial is 8x²-18x-9

Step-by-step explanation:

If $\alpha$ and $\beta$ are the zeroes of quadratic polynomial

$f(x)=x^2-3x-2$ then

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

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

Also,

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

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

We have to find the quadratic polynomial whose zeroes are $\alpha+\frac{1}{2\beta}$ and $\beta+\frac{1}{2\alpha}$

Sum of zeroes

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

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

$=3+\frac{\alpha+\beta}{2\alpha\beta}$

$=3+\frac{3}{2\times(-2)}$

$=3-\frac{3}{4}$

$=\frac{9}{4}$

Product of zeroes

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

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

$=-2+1+\frac{1}{4\times(-2)}$

$=-1-\frac{1}{8}$

$=-\frac{9}{8}$

We know that quadratic polynomial whose zeroes are $\alpha$ and $\beta$ is given by

$x^2-(\alpha+\beta)x+\alpha\beta$

Therefore, the quadratic polynomial is

$x^2-(\frac{9}{4})x+(-\frac{9}{8})$

or, $x^2-\frac{18}{8}x-\frac{9}{8}$

or, $8x^2-18x-9$

Hope this answer is helpful.

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