$\alpha$ and $\beta$ are real roots of $p^2x^2+2x+p=0$, where $0<p\leq1$. Find the quadratic with roots $1/\alpha+1$ and $1/\beta+1$, in terms of $p$.

Question

$\alpha$ 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?

Answer 1

You can use the given identities

Given a quadratic equation $ax^2+bx+c$,

• Sum of roots is $-b/a$
• Product of roots is $c/a$

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.