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
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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.
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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.
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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 \(-17, 17\).