If $\alpha$ and $\beta$ are zeros of the polynomial $4x^2 -x -4$, find the quadratic whose zeros are $\frac{1}{2\alpha}$ and $\frac{1}{2\beta}$.

Question

I am utterly stuck at this question which I am sure has a simple solution, but I just can't seem to be able to see it. Any help would be greatly appreciated.

If $\alpha$ and $\beta$ are zeros of the polynomial $4x^2 -x -4$, find the quadratic polynomial whose zeros are $\frac{1}{2\alpha}$ and $\frac{1}{2\beta}$.

Thank you!

Answer 1

Consider $$\left(x-\frac{1}{2 \alpha }\right) \left(x-\frac{1}{2 \beta }\right)=x^2-\frac 12 \left(\frac{1}{ \alpha }+\frac{1}{ \beta }\right)x+\frac{1}{4 \alpha \beta }=x^2-\frac 12 \left(\frac{\alpha +\beta}{ \alpha\beta } \right)x+\frac{1}{4 \alpha \beta }$$

Answer 2

Let $y=\frac{1}{2x}$. Then $4y^2-y-4$ cancels when $y=\alpha$ or $y=\beta$. But $$y=\alpha \Leftrightarrow \frac{1}{2x}=\alpha \Leftrightarrow x= \frac{1}{2\alpha}.$$ Now all you have to do is express $4y^2-y-4$ in terms of $x$. You will get a rational fraction, but it can only cancel when its numerator does; hence this numerator is your polynomial.

Answer 3

If $$4x^2-x-4=0$$ is true for $x=\alpha,\beta$, then $$4(2y)^{-2}-(2y)^{-1}-4=0$$ holds for $y=1/2\alpha,1/2\beta$.

You should be able to conclude.

Answer 4

You have $4x^2-x-4 = (x-\alpha)(x-\beta)$. If you multiply the last bit out, you get expressions for $\alpha+\beta$ and $\alpha\beta$. The quadratic you're looking for is $(x-1/2\alpha)(x-1/2\beta)$. If you multiply this out and clear fractions, the expressions you found above should give you your answer.