Let $x_1,x_2 \in \mathbb{R}$ be the roots of the equation $x^2+px+q=0$

Question

Let $x_1,x_2 \in \mathbb{R}$ be the roots of the equation $x^2+px+q=0$. Find $p$ and $q$ if it is known that $x_1+1$ and $x_2+1$ are the roots of the equation $x^2-p^2x+pq=0$.

The roots of $x^2+px+q=0$ satisfy $$x_1+x_2=-\dfrac{b}{a}=-p,x_1x_2=\dfrac{c}{a}=q.$$ Also the roots of $x^2-p^2x+pq=0$ satisfy $$x_1+x_2=x_1+1+x_2+1=x_1+x_2+2=-\dfrac{b}{a}=p^2,x_1x_2=(x_1+1)(x_2+1)=\dfrac{c}{a}=pq.$$ Is it confusing I am using $x_1$ and $x_2$ for the roots of both equations? How can I improve what I've written? What to do next?

Answer 1

Next substitute $x_1+x_2=-p$ into $x_1+x_2+2=p^2$ to obtain $2-p=p^2$, which you can solve for $p$.

To find $q$, expand $(x_1+1)(x_2+1)=pq$ to get $x_1x_2+(x_1+x_2)+1=pq$, which is the same as $q-p+1=pq$.

And you should not use $x_1,x_2$ to mean two different things.

Answer 2

Another way :

Like If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.

let $x_k+1=y_k; k=1,2$

$\implies x_k=y_k-1$

As $x_k$ is a root of $$x^2+px+q=0$$

$$(y_k-1)^2+p(y_k-1)+q=0$$

Answer 3

Well, I tried this:

Set the equations:

$-p = x_1 + x_2 (a), x_1x_2 = q (b) $

$p^2 = x_1 +x_2 +2 (c), (x_1 + 1)(x_2 + 1) = pq (d)$

So, let's substitute (a) in (c), then we will have:

$p^2 = -p + 2$

Solving it, we'll have:

$p=-2$

$p=1$

About $q$:

Let's expand the product of (d) and substitute (b) in it. Then, we'll have:

$q -p +1=pq$

Since we have the equation (d,b), we have to check for which values of $p$ it still makes sense. Therefore:

$(p,q)={(-2,-1),(1,-1)}$