What is the smallest value of $x$ that satisfies the equation $8x^2 - 38x + 35 = 0$? Express your answer as a decimal.
We see that we can rewrite the left side of the equation $8x^2 - 38x + 35$ as $(2x - 7)(4x - 5)$, so we have $(2x - 7)(4x - 5) = 0$. Thus, solving the equations $2x - 7 = 0$ and $4x - 5 = 0$ gives us $x = 3.5$ and $x = 1.25$ as our solutions. Since $1.25 < 3.5$, our final answer is $x = \boxed{1.25}$.
1.25
Level 3
Algebra
Problem: What is the smallest value of $x$ that satisfies the equation $8x^2 - 38x + 35 = 0$? Express your answer as a decimal.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the smallest value of \( x \) that satisfies the quadratic equation \( 8x^2 - 38x + 35 = 0 \), we can apply the quadratic formula, which is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 8 \), \( b = -38 \), and \( c = 35 \).

First, we calculate the discriminant \( b^2 - 4ac \):
1. Compute \( b^2 \): 
   \[
   (-38)^2 = 1444
   \]
2. Compute \( 4ac \):
   \[
   4 \cdot 8 \cdot 35 = 1120
   \]
3. Now, find the discriminant:
   \[
   1444 - 1120 = 324
   \]

Next, we take the square root of the discriminant:
\[
\sqrt{324} = 18
\]

Now, we substitute \( b \) and the square root of the discriminant back into the quadratic formula:
\[
x = \frac{-(-38) \pm 18}{2 \cdot 8} = \frac{38 \pm 18}{16}
\]

This gives us two potential solutions:
1. For the positive case:
   \[
   x = \frac{38 + 18}{16} = \frac{56}{16} = 3.5
   \]
2. For the negative case:
   \[
   x = \frac{38 - 18}{16} = \frac{20}{16} = 1.25
   \]

Among the two solutions, \( 1.25 \) is the smaller value. Therefore, the smallest value of \( x \) that satisfies the equation is \( 1.25 \).<|END_OF_LATENT|> The final answer is $1.25$.