Question

x
-23
0
12
x^2+20x=69
Determine the solution and identify which of these values is in the solution
Only x= __ is in the solution set

Answer 1

To solve the equation \( x^2 + 20x = 69 \), we first rearrange it into standard quadratic form:

\[
x^2 + 20x - 69 = 0
\]

Next, we can solve for \( x \) using the quadratic formula:

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

where \( a = 1 \), \( b = 20 \), and \( c = -69 \).

Calculating the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 20^2 - 4 \cdot 1 \cdot (-69) = 400 + 276 = 676
\]

Now substituting the values into the quadratic formula:

\[
x = \frac{-20 \pm \sqrt{676}}{2 \cdot 1}
\]

Since \( \sqrt{676} = 26 \):

\[
x = \frac{-20 \pm 26}{2}
\]

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

1. Using the plus sign:

\[
x = \frac{-20 + 26}{2} = \frac{6}{2} = 3
\]

2. Using the minus sign:

\[
x = \frac{-20 - 26}{2} = \frac{-46}{2} = -23
\]

So, the solutions to the equation \( x^2 + 20x = 69 \) are \( x = 3 \) and \( x = -23 \).

Now, let's identify which of the given values (-23, 0, 12) is in the solution set. The solution set is \{3, -23\}.

The only value from the list that is in the solution set is:

\[
\text{Only } x = -23 \text{ is in the solution set.}
\]