B4 Let $a$ be the largest real value of $x$ for which $x^{3}-8 x^{2}-2 x+3=0$. Determine the integer closest to $a^{2}$.
Solution 1
Since the equation has degree 3 , there are at most 3 values of $x$ for which it will hold.
Let $f(x)=x^{3}-8 x^{2}-x+3$, and $b, c$ the other two roots of $f(x)$.
Note that
$$
f(-1)=(-1)^{3}-8(-1)^{2}-2(-1)+3=-40 .
$$

Hence, there is a root between -1 and $-1 / 2$.
Similarly,
$$
f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{3}-8\left(\frac{1}{2}\right)^{2}-2 \cdot \frac{1}{2}+3=\frac{1}{8}-2-1+3=\frac{1}{8}>0
$$
and
$$
f(1)=1-8-2+3=-61 / 2, b^{2}+c^{2}>1 / 2$. Now we need an upper bound on $b^{2}+c^{2}$. Note that
$$
f\left(\frac{1}{\sqrt{2}}\right)=\left(\frac{1}{\sqrt{2}}\right)^{3}-8\left(\frac{1}{\sqrt{2}}\right)^{2}-2\left(\frac{1}{\sqrt{2}}\right)+3=\frac{1}{2 \sqrt{2}}-4-\frac{2}{\sqrt{2}}+3=\frac{-3}{2 \sqrt{2}}-10,1 / 2<c<1 / \sqrt{2}$.
Therefore, $b^{2}+c^{2}<1+1 / 2=3 / 2$. Since $1 / 2<b^{2}+c^{2}<3 / 2$ and $a^{2}+b^{2}+c^{2}=68$, $66.5<a^{2}<67.5$. Therefore, the integer closest to $a^{2}$ is 67 .
Solution 2
As in solution 1 , we can verify that there are two values of $x$ between -1 and 1 for which the equation holds. Note that since the equation is cubic there are at most 3 distinct solutions.
We can rewrite the equation as $x^{2}(x-8)=2 x-3$, which simplifies to $x^{2}=2+\frac{13}{x-8}$. Letting $x=8.2$ we get the left hand side is $8.2^{2}=67.24$ and the right side is $2+\frac{13}{2}=67$. As we decrease $x$, from 8.2 to 8.1 , the left hand side decreases from 67.24 to 65.61 and the right hand side increases from 67 to 132 . Since both functions are continuous, there is a point between where they will have the same value, and that value will be between 67 and 67.24. Thus, the integer closest to $x^{2}$ is 67 .