Solution Sets of Quadratic Equations Practice

Question

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2 of 52 of 5 Items

Question
Use the table to answer the question.

x Substituted Evaluate True Statement?
−23

0
12
For the quadratic equation x2+20x=69
, complete the table by testing each of the given values to determine whether it is a solution. Identify which one of these values is in the solution set.

(1 point)
Only x=
is in the solution set.

Answer 1

To determine which value is in the solution set for the quadratic equation \( x^2 + 20x = 69 \), we will first set the equation to equal zero by rearranging it:

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

Next, we'll evaluate the equation for each of the given \( x \) values in the table: \( -23 \), \( 0 \), and \( 12 \).

1. For \( x = -23 \): \[ (-23)^2 + 20(-23) - 69 = 529 - 460 - 69 = 0 \] True Statement? Yes, it equals 0.

2. For \( x = 0 \): \[ (0)^2 + 20(0) - 69 = 0 + 0 - 69 = -69 \] True Statement? No, it does not equal 0.

3. For \( x = 12 \): \[ (12)^2 + 20(12) - 69 = 144 + 240 - 69 = 315 \] True Statement? No, it does not equal 0.

After evaluating all the values, we conclude:

• \( -23 \) is in the solution set since it satisfies the equation \( x^2 + 20x - 69 = 0 \).
• The values \( 0 \) and \( 12 \) do not satisfy the equation.

Therefore:

Only \( x = -23 \) is in the solution set.