sempteim245

2022-03-30

Prove an inequality involving a root of a quadratic equation
If $x=\rho$
is a solution to: ${x}^{2}+bx+c=0$
Prove that $|\rho |-1<|b|+|c|$

### Answer & Explanation

Tristatex9tw

Step 1
Let ${\rho }_{1}$ and ${\rho }_{2}$ be the solutions to ${x}^{2}+bx+c=0$/
Let's assume, without loss of generality, that $|{\rho }_{1}\ge |{\rho }_{2}\mid$
We shall consider two cases:
a) If $|{\rho }_{1}|<1$, then
$|{\rho }_{2}|-1\le |{\rho }_{1}|-1<0\le |b|+|c|$
b) If $|{\rho }_{1}|\ge 1$, it follows from Vieta's formulas and the reverse triangular inequality that
$|b|+|c|=|{\rho }_{1}+{\rho }_{2}|+|{\rho }_{1}{\rho }_{2}|\ge ||{\rho }_{1}|-|{\rho }_{2}||+|{\rho }_{1}||{\rho }_{2}|$
$=|{\rho }_{1}|-|{\rho }_{2}|+|{\rho }_{1}||{\rho }_{2}|$
$=\left(|{\rho }_{1}|-1\right)\left(|{\rho }_{2}|+1\right)+1>\left(|{\rho }_{1}|-1\right)\left(|{\rho }_{2}|+1\right)\ge |{\rho }_{1}|-1\ge |{\rho }_{2}|-1.$
Therefore, in both cases it is true that $|\rho |-1<|b|+|c|$

microsgopx6z7

Step 1
Consider the matrix
$A=\left[\begin{array}{cc}0& 1\\ -c& -b\end{array}\right]$
then ${x}^{2}+bx+c$ is the corresponding characteristic polynomial.
By Gershgorin's theorem, we have $|\rho |\le 1$ or
$|\rho +b|\le |c|$
By reversed triangle inequality, $|\rho |\le 1$ or $|\rho |\le |c|+|b|$,
hence $|\rho |\le max\left(1,|b|+|c|\right)\le 1+|b|+|c|$

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