It is given the equation $x^2+px+1=0$, with roots $x_1$ and $x_2$; (a) find a second-degree equation with roots $y_1,y_2$ satisfying the conditions $y_1=x_1(1-x_1)$, $y_2=x_2(1-x_2)$; (b) find all possible values of the real parameter $p$ such that the roots of the new equation lies between $-2$ and $1$.
Problem
Source:
Tags: algebra, Polynomials
TheMathematics
24.06.2021 08:33
b) vieta formula $ \rightarrow x^2-Sx+q=0 \rightarrow S=x_1 + x_2$ & $ q=x_1 \times x_2 $ $x_1 \times x_2 =1 \rightarrow x_2=\frac{1}{x_1}$ then $x_1+\frac{1}{x_1} = [-2,1]$ and we know $x+\frac{1}{x} \ge 2 $ for positive number and $x+\frac{1}{x} \le -2 $ for negative number. It's only when the equality occurs $x=\frac{1}{x} \rightarrow x=\frac{1}{x}= -1$ or $x=\frac{1}{x}=1$ so p= 2 or -2. a) for $x_1=\frac{1}{x_1}=1$ $y_1=y_2=0 \rightarrow x^2=0$ and for $x_1=\frac{1}{x_1}=-1$ $y_1=y_2=-2 \rightarrow x^2+4x+4=0$
natmath
24.06.2021 08:41
By viete's
$$y_1+y_2=x_1+x_2-(x_1^2+x_2^2)$$$$y_1+y_2=x_1+x_2-(x_1+x_2)^2+2x_1x_2$$$$y_1+y_2=-p-p^2+2$$
We also have
$$y_1y_2=(0-x_1)(0-x_2)(1-x_1)(1-x_2)$$$$y_1y_2=(1)(2+p)=p+2$$
So the monic polynomial with roots $y_1,y_2$ is
$$f(x)=x^2+(p^2+p-2)x+p+2$$
We require that $y_1,y_2\in (-2,1)$. Consider the solutions to
$$-2<x(1-x)<1$$This is true when $x\in (-1,2)$ or when $Re(x)=\frac{1}{2}$ and $|Im(x)|<\frac{\sqrt{3}}{2}$.
Clearly both $x_1$ and $x_2$ must both satisfy the same one of these conditions since $p$ is real.
Case 1: $x_1,x_2\in\mathbb{R}$.
This means they are both in the interval $(-1,2)$
Since $x_1x_2=1$, it is clear that both $x_1$ and $x_2$ must be positive unlessthe question intended that $y_1$ and $y_2$ can be $-2$ or $1$, which would then mean that a possible value of $p$ would be $p=2$
We want to find the range of $p=-x_1-x_2$ given that $x_1x_2=1$ and $x_1,x_2\in (0,2)$.
WLOG $x_2\geq x_1$, which implies $x_2\geq 1$. Since $p=-x_2-\frac{1}{x_2}$ is monotonic, it is clear that the range of $p$ is $\boxed{(-\frac{5}{2},-2]}$
Case 2: $x_1,x_2\not\in\mathbb{R}$
This means that the real parts of $x_1,x_2$ are $\frac{1}{2}$. Since they must be conjugates, the only value of $p$ is $-1$.
So $p$ must lie in $\boxed{(-\frac{5}{2},-2]\cup\{-1\}}$ or depending on the wording of the question$\boxed{[-\frac{5}{2},-2]\cup\{-1,2\}}$
It's also not too bad for part b) if you just use intermediate value theorem and properties of quadratics with just the equation we got from a). I started using that method and you get a quartic, so decided to not do that. However, it turns out the quartic is easily factorable.
DAVROS
24.06.2021 11:22
(a) $y=x(1-x) \implies x^2-x+y=0 \implies (p+1)x=y-1.\:$ If $p=-1$ then $y+1=0$ which is not quadratic. If $p\ne-1$ then back substitution gives required quadratic as $f(y) = y^2+(p^2+p-2)y+2+p=0$ (b) Three conditions apply : $f(y)=0$ has real roots : $f(-2)>0 : f(1)>0$ Because of $(p+1)x=y-1, \text{(y is real)} \iff \text{(x is real)}$ so the first condition is $p^2\ge 4$ The other conditions give $ -2p^2-p+10>0$ and $(p+1)^2>0$ which imply $-\frac 52 < p < 2$ Combining these : $-\frac 52 <p \le-2,$