What is the largest value of $x$ such that the expression \[\dfrac{x+1}{8x^2-65x+8}\] is not defined?
In this particular case, the fraction will be undefined only if its denominator is equal to zero. Because of this, we can ignore the numerator. We start by setting the binomial in the denominator equal to 0: \begin{align*} 8x^2-65x+8=0
\\\Rightarrow\qquad (8x-1)(x-8)=0
\end{align*} We find that the two possible values for $x$ are $\frac18$ and $8$. Since the question asks for the largest value, the final solution is $\boxed{8}$.
8
Level 3
Algebra
Problem: What is the largest value of $x$ such that the expression \[\dfrac{x+1}{8x^2-65x+8}\] is not defined?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine when the expression is not defined, we need to identify when the denominator equals zero. The denominator is given by the quadratic expression \(8x^2 - 65x + 8\). 

First, we set the denominator equal to zero:

\[ 8x^2 - 65x + 8 = 0. \]

Next, we apply the quadratic formula, which is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]

where \(a = 8\), \(b = -65\), and \(c = 8\). 

Calculating the discriminant:

\[ b^2 - 4ac = (-65)^2 - 4 \cdot 8 \cdot 8 = 4225 - 256 = 3969. \]

Now, we find the square root of the discriminant:

\[ \sqrt{3969} = 63. \]

Substituting back into the quadratic formula gives:

\[ x = \frac{65 \pm 63}{16}. \]

Calculating the two possible values for \(x\):

1. For the positive root:

\[ x = \frac{65 + 63}{16} = \frac{128}{16} = 8. \]

2. For the negative root:

\[ x = \frac{65 - 63}{16} = \frac{2}{16} = \frac{1}{8}. \]

The expression is not defined at these two values of \(x\). Among these, the largest value is \(8\). 

Thus, the largest value of \(x\) such that the expression is not defined is \(8\).<|END_OF_LATENT|> The final answer is $8$.