Question 1
To determine which of the values is in the solution set for the quadratic equation \( -2x^2 - 3x + 2 = 0 \), we need to substitute each value of \( x \) into the equation and evaluate it.
Let's substitute each value:
For \( x = -34 \):
\[ -2(-34)^2 - 3(-34) + 2 = -2(1156) + 102 + 2 = -2312 + 102 + 2 = -2210 \quad (\text{Not } 0) \]
For \( x = -12 \):
\[ -2(-12)^2 - 3(-12) + 2 = -2(144) + 36 + 2 = -288 + 36 + 2 = -250 \quad (\text{Not } 0) \]
For \( x = 12 \):
\[ -2(12)^2 - 3(12) + 2 = -2(144) - 36 + 2 = -288 - 36 + 2 = -322 \quad (\text{Not } 0) \]
None of the values provided yield a true statement (i.e., evaluate to zero). Therefore: Only \( x = \text{none of these values} \) is in the solution set.
Question 2
To solve the equation \( x^2 = 289 \) by inspection:
We can find \( x \) by taking the square root of both sides: \[ x = \pm \sqrt{289} = \pm 17 \]
The two solutions are \( -17 \) and \( 17 \). Listing the lesser number first: { -17, 17 }
Question 3
To solve the equation \( -9 + (x - 3)^2 - 4 = -4 \):
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Simplify the equation: \[ -9 + (x - 3)^2 - 4 = -4 \ (x - 3)^2 - 13 = -4 \quad \text{(adding 13 to both sides)} \ (x - 3)^2 = 9 \]
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Taking the square root of both sides: \[ x - 3 = \pm 3 \]
So we have two cases:
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Case 1: \( x - 3 = 3 \) \ \( x = 6 \)
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Case 2: \( x - 3 = -3 \) \ \( x = 0 \)
Thus, the values in the solution set are: { 0, 6 }
From the provided responses, the values that are included in the solution set would be listed among the responses, which are not complete. Based on what we've calculated:
- 0 (not listed)
- 6 (not listed)
If given values include the correct integers with one of them as the answer, ensure that you compare between \( 0 \) and \( 6 \) if those numbers exist in your options provided.
Please verify against your list for exact matching values.