Question

For the quadratic equation-2x^3 -3x + 2 =0 complete the table
x Substituted Evaluate True
-3/4 ? ? ?
-1/2 ? ? ?
1/2 ? ? ?
Identify which one of the values is in the solution set only x = ____ is in the solution set

Solve x^2 = 289 there are two real solutions. Enter the lesser number first and the greater number last

Answer 1

To complete the table for the cubic equation \(-2x^3 - 3x + 2 = 0\), we'll substitute each \(x\) value into the equation, evaluate it, and determine if it is true (i.e., if the expression equals zero).

### Step 1: Substitute and Evaluate
1. **For \(x = -\frac{3}{4}\)**:
\[
-2\left(-\frac{3}{4}\right)^3 - 3\left(-\frac{3}{4}\right) + 2
\]
Calculating \(\left(-\frac{3}{4}\right)^3 = -\frac{27}{64}\):
\[
-2\left(-\frac{27}{64}\right) + \frac{9}{4} + 2 = \frac{54}{64} + \frac{9}{4} + 2
\]
Convert \(\frac{9}{4}\) and \(2\) to have a common denominator:
\[
\frac{9}{4} = \frac{144}{64}, \quad 2 = \frac{128}{64}
\]
Now add it all together:
\[
\frac{54 + 144 + 128}{64} = \frac{326}{64} \neq 0
\]
False.

2. **For \(x = -\frac{1}{2}\)**:
\[
-2\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right) + 2
\]
Calculating \(\left(-\frac{1}{2}\right)^3 = -\frac{1}{8}\):
\[
-2\left(-\frac{1}{8}\right) + \frac{3}{2} + 2 = \frac{1}{4} + \frac{3}{2} + 2
\]
Convert \(\frac{3}{2}\) and \(2\):
\[
\frac{3}{2} = \frac{6}{4}, \quad 2 = \frac{8}{4}
\]
Now add:
\[
\frac{1 + 6 + 8}{4} = \frac{15}{4} \neq 0
\]
False.

3. **For \(x = \frac{1}{2}\)**:
\[
-2\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) + 2
\]
Calculating \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\):
\[
-2\left(\frac{1}{8}\right) - \frac{3}{2} + 2 = -\frac{1}{4} - \frac{3}{2} + 2
\]
Convert \(-\frac{3}{2}\) and \(2\):
\[
-\frac{3}{2} = -\frac{6}{4}, \quad 2 = \frac{8}{4}
\]
Now add:
\[
-\frac{1}{4} - \frac{6}{4} + \frac{8}{4} = \frac{1}{4} \neq 0
\]
False.

### Summary of the Table
| \(x\) | Substituted | Evaluate | True |
|-------------|-----------------------------------|-----------------------|--------|
| \(-\frac{3}{4}\) | \(-2\left(-\frac{3}{4}\right)^3 - 3\left(-\frac{3}{4}\right) + 2\) | \(\frac{326}{64}\) | False |
| \(-\frac{1}{2}\) | \(-2\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right) + 2\) | \(\frac{15}{4}\) | False |
| \(\frac{1}{2}\) | \(-2\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) + 2\) | \(\frac{1}{4}\) | False |

**Conclusion**: None of the values provided are in the solution set.

### Step 2: Solve \(x^2 = 289\)
To solve \(x^2 = 289\), we take the square root of both sides:
\[
x = \pm \sqrt{289}
\]
Calculating:
\[
\sqrt{289} = 17
\]
Thus, the two real solutions are:
\[
x = -17 \quad \text{and} \quad x = 17
\]
**Final Answer**: The solutions are \(-17, 17\).