Question

Solve the factored quadratic equation $(19x-3)(9+x)=0$
The solution set is $x=\{ -9,\frac {19}{3}\}$
The solution set is $x=\{ -9,\frac {3}{10}\}$
The solution set is $x=\{ \frac {19}{3},9\}$
The solution set is $x=\{ -9,3\} $

Answer 1

To solve the factored quadratic equation \((19x-3)(9+x) = 0\), we need to find the values of \(x\) that make either of the factors equal to zero.

1. **Set the first factor to zero:**
\[
19x - 3 = 0
\]
Solving for \(x\), we add \(3\) to both sides:
\[
19x = 3
\]
Now divide by \(19\):
\[
x = \frac{3}{19}
\]

2. **Set the second factor to zero:**
\[
9 + x = 0
\]
Solving for \(x\), we subtract \(9\) from both sides:
\[
x = -9
\]

Thus, the solution set for the equation \((19x-3)(9+x) = 0\) is:
\[
x = \{-9, \frac{3}{19}\}
\]

Now, we can check the provided options against the solutions we derived.

- The solution set is \(x=\{ -9,\frac {19}{3}\}\)
- The solution set is \(x=\{ -9,\frac {3}{10}\}\)
- The solution set is \(x=\{ \frac {19}{3},9\}\)
- The solution set is \(x=\{ -9,3\}\)

Among these, \( \frac{19}{3} \) and \( \frac{3}{10} \) are not solutions based on our calculations, nor is \(9\) or \(3\). Therefore, the provided solution set is incorrect.

Final conclusion:
The correct solution set is \(x = \{-9, \frac{3}{19}\}\).