[Geometry]
Problem: Let $T=23$. In triangle $A B C$, the altitude from $A$ to $\overline{B C}$ has length $\sqrt{T}, A B=A C$, and $B C=T-K$, where $K$ is the real root of the equation $x^{3}-8 x^{2}-8 x-9=0$. Compute the length $A B$.

Solution: Rewrite the equation as $x^{3}-1=8\left(x^{2}+x+1\right)$, so that $(x-1)\left(x^{2}+x+1\right)=8\left(x^{2}+x+1\right)$. Because $x^{2}+x+1$ has no real zeros, it can be canceled from both sides of the equation to obtain $x-1=8$ or $x=9$. Hence $B C=T-9$, and $A B^{2}=(\sqrt{T})^{2}+\left(\frac{T-9}{2}\right)^{2}=T+\left(\frac{T-9}{2}\right)^{2}$. Substitute $T=23$ to obtain $A B=\sqrt{72}=\mathbf{6} \sqrt{\mathbf{2}}$.

Final answer: $6 \sqrt{2}$