$$x^{3}-2bx^{2}-a^{2}x+b^{2}=0$$ 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. :) $\endgroup$– MaximeJacconCommented Feb 29 at 8:42
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$\begingroup$ @MaximeJaccon You can only use the rational root theorem if the coefficients are integers $\endgroup$– MiltenCommented Feb 29 at 9:32
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1 Answer
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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?