To determine which values are solutions to the quadratic equation \( z^2 + 20x = 69 \), we need to substitute each given value of \( x \) into the equation and evaluate whether the equation holds true.
First, let's rearrange the equation to express \( z^2 \):
\[ z^2 = 69 - 20x \]
Now we'll substitute each value of \( x \) into this equation:
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For \( x = -23 \): \[ z^2 = 69 - 20(-23) = 69 + 460 = 529 \] \[ z = \pm \sqrt{529} = \pm 23 \] This is a valid solution.
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For \( x = 0 \): \[ z^2 = 69 - 20(0) = 69 \] \[ z = \pm \sqrt{69} \quad (\text{This is a real number, but not an integer.}) \] This is a valid solution but not an integer.
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For \( x = 12 \): \[ z^2 = 69 - 20(12) = 69 - 240 = -171 \] Since \( z^2 \) cannot be negative (in the real number domain), this value does not produce a valid solution.
Now let's fill in the table:
| X | Substituted | Evaluate | True Statement? | |-----|---------------------------|----------------------|-------------------------------| | -23 | \( z^2 = 69 - 20(-23) \) | \( z^2 = 529 \) | True (valid solutions: ±23) | | 0 | \( z^2 = 69 - 20(0) \) | \( z^2 = 69 \) | True (valid solutions: ±√69) | | 12 | \( z^2 = 69 - 20(12) \) | \( z^2 = -171 \) | False (no real solutions) |
From the evaluation, we conclude:
- The value \( -23 \) produces integer solutions.
- The value \( 0 \) produces real solutions.
- The value \( 12 \) does not produce any real solutions.
Thus, it can be said that:
Only \( x = -23 \) is in the solution set.