Question

Use the table to complete the statement.

x Substituted Evaluate True Statement?
−34

−12

12

Answer 1

For the quadratic equation −2x2−3x+2=0
, complete the table by testing each of the given values to determine whether it is a solution. Identify which one of the values is in the solution set. Only x =

is in the solution set.

Answer 2

Solve x2=289
by inspection. There are two real solutions. Enter the lesser number first.

Answer 3

It seems you've provided a table structure but not the actual entries in the "Substituted," "Evaluate," and "True Statement?" columns corresponding to each value of \( x \). To help you complete the statement based on that table, I would need more information on what computations or evaluations you're trying to perform for each of the values of \( x \):

- What expression or formula is being substituted for each \( x \)?
- What is being evaluated to determine if the statement is true?

Please provide additional details so I can assist you accurately!

Answer 4

To determine if the values \( x = -34, -12, \) and \( 12 \) are solutions to the quadratic equation \(-2x^2 - 3x + 2 = 0\), we will substitute each value into the equation and evaluate.

Here’s how to do it step by step for each value:

1. **For \( x = -34 \):**
\[
-2(-34)^2 - 3(-34) + 2
= -2(1156) + 102 + 2
= -2312 + 102 + 2
= -2312 + 104
= -2208 \quad (\text{Not } 0)
\]

2. **For \( x = -12 \):**
\[
-2(-12)^2 - 3(-12) + 2
= -2(144) + 36 + 2
= -288 + 36 + 2
= -288 + 38
= -250 \quad (\text{Not } 0)
\]

3. **For \( x = 12 \):**
\[
-2(12)^2 - 3(12) + 2
= -2(144) - 36 + 2
= -288 - 36 + 2
= -288 - 34
= -322 \quad (\text{Not } 0)
\]

### Summary of Evaluations:

- For \( x = -34 \): evaluation resulted in \(-2208\)
- For \( x = -12 \): evaluation resulted in \(-250\)
- For \( x = 12 \): evaluation resulted in \(-322\)

### Conclusion:
None of the values provided (−34, −12, or 12) satisfy the equation \(-2x^2 - 3x + 2 = 0\) since they do not produce a result of 0.

Therefore, the statement should be:
**Only \( x = \) (none of the options provided) is in the solution set.** In other words, none of the given values are in the solution set of the equation.

Answer 5

To solve the equation \( x^2 = 289 \) by inspection, we need to find the values of \( x \) that satisfy this equation.

We can take the square root of both sides:

\[
x = \pm \sqrt{289}
\]

Calculating the square root:

\[
\sqrt{289} = 17
\]

Thus, the solutions are:

\[
x = 17 \quad \text{and} \quad x = -17
\]

When we list the solutions with the lesser number first, we have:

\[
-17, \ 17
\]

So, the two real solutions are \( -17 \) and \( 17 \).