Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $x^2+8x+6$, form a quadratic polynomial whose zeros are $\frac{1}{\alpha }$ and $\frac{1}{\beta }$.

Answer 1

$x^2+8x+6=0$

Sum $=\alpha +\beta =-8$

Product $=\alpha \beta =6$

Now equation whose zeroes are $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ is $(x-\frac{1}{\alpha })(x-\frac{1}{\beta })$

$x^2–x(\frac{1}{\alpha }+\frac{1}{\beta })+\frac{1}{\alpha \beta }$

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

$=x^2–x\left(\frac{-8}{6}\right)+\frac{1}{6}$

$=x^2+\frac{4x}{3}+\frac{1}{6}$