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$$x^{3}-2bx^{2}-a^{2}x+b^{2}=0$$ Show that if $x=-1$ is a solution, then $1-\sqrt{2}\le b\le1+\sqrt{2}$


I subbed in the solution $x=-1$, completed the square, and now I'm left with the equation $\left(a-1\right)^{2}+b^{2}=2$

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  • $\begingroup$ Can't you just use rational root theorem, thus $b^2$ has to be $\pm1$, and if b is real, b must be 1, which satisfies the inequality. :) $\endgroup$ Feb 29 at 8:42
  • $\begingroup$ @MaximeJaccon You can only use the rational root theorem if the coefficients are integers $\endgroup$
    – Milten
    Feb 29 at 9:32

1 Answer 1

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You seem to have made a mistake during simplification and/or completing the square.

Substituting $x = -1$ into the given equation, we get

$$\begin{align*} -1 - 2b + a^2 + b^2 &= 0 \\[0.3cm] \Rightarrow \;\;\;\;\;\;a^2 + (b-1)^2 &= 2 \end{align*}$$

Since $a^2 \ge 0$,

$$\begin{align*} (b-1)^2 &\le 2 \\[0.3cm] \end{align*}$$

Can you take it from here?

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