$\alpha$ and $\beta$ are roots of the equation $p^2x^2 + 2x + p = 0$, where $0<p\leq1$, and $\alpha, \beta$ are real numbers. Find the quadratic equation whose roots are $1/\alpha+1$ and $1/\beta+1$, in terms of $p$.
I tried setting
$$p^2x^2 + 2x + p = (x - \alpha)(x - \beta)$$
Then solved to get:
$$p^2x^2 + 2x + p = x^2 - \alpha x -\beta x + \alpha \beta$$
So that means $2 = - \alpha -\beta$, $p = αβ$, and $p^2 = 1$.
Is that correct?
On
Well here is the hint Suppose $y=\frac{1}{1+a}$ Now solve $a$ in terms of $y$ Now $a$ is root of the equation so $a$ in terms $y$ is a root of the equation so we know $a$ plugs upto 0 yes, you are right you need to make a quadratic in $y$ now and Congratulations you are done!!! Do it quickly!!!
On
One simple way of working this is to just... solve the given equation. No need to do anything fancy, just the good ol' quadratic formula. If we do, we get
$$ \alpha, \beta = \frac{-1 \pm \sqrt{1-p^3}}{p^2}$$
We can assume WLOG that $\alpha$ is the larger root. Then find $\alpha' = 1/\alpha +1, \beta' = 1/\beta +1$ from those:
$$\alpha' = 1+\frac{1}{\alpha} = \frac{p^2-1 + \sqrt{1-p^3}}{-1 + \sqrt{1-p^3}}$$
$$\beta' = \frac{p^2-1 - \sqrt{1-p^3}}{-1 - \sqrt{1-p^3}}$$
Note: These equations have been edited from the original ones, which were very, very wrong.
Now you can take those, and substitute them into $(x-\alpha')(x-\beta')=0$. A bit of algebra will get you to your final polynomial.
If you meant to write $1/(\alpha +1), 1/(\beta +1)$, then you'll need to find those instead, but the rest is the same.
Good luck with the rest, I hope to see the final answer from you soon!
You can use the given identities
Given a quadratic equation $ax^2+bx+c$,
Hint: Reduce $\frac{1}{\alpha+1}+\frac{1}{\beta+1}$, and $\frac{1}{\alpha+1}\cdot\frac{1}{\beta+1}$. Note that you know $\alpha+\beta$, and $\alpha\cdot\beta$ from above.
You can construct the quadratic equation from sum, product of roots, using the above points.