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or the quadratic equation −2x2−3x+2=0 , complete the table by testing each of the given values to determine whether it is a sol...Question
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.
, 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.
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Answered by
GPT-4o mini
To determine which values solve the quadratic equation \(-2x^2 - 3x + 2 = 0\), we need to evaluate the equation for each given value of \(x\) and see if it results in zero.
Let's first rewrite the equation:
\[
-2x^2 - 3x + 2 = 0
\]
We will test each proposed value in the equation. Here are the steps to check if each value is a solution:
1. Substitute the value of \(x\) into the left-hand side of the quadratic equation.
2. Evaluate the expression to see if it equals zero.
Let’s test the potential values (you'll need to provide the specific values you're checking if they are not provided):
### Example Process:
1. **Let’s assume we want to test the values.** For instance, let's say we want to test \(x = 1\):
\[
-2(1)^2 - 3(1) + 2 = -2(1) - 3 + 2 = -2 - 3 + 2 = -3 \quad (\text{not a solution})
\]
2. **Now let’s test \(x = -1\):**
\[
-2(-1)^2 - 3(-1) + 2 = -2(1) + 3 + 2 = -2 + 3 + 2 = 3 \quad (\text{not a solution})
\]
3. **Now testing \(x = \frac{1}{2}\):**
\[
-2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2 = -\frac{1}{2} - \frac{3}{2} + 2 = -2 + 2 = 0 \quad (\text{is a solution})
\]
4. **And testing \(x = 2\):**
\[
-2(2)^2 - 3(2) + 2 = -2(4) - 6 + 2 = -8 - 6 + 2 = -12 \quad (\text{not a solution})
\]
5. **Now testing \(x = -2\):**
\[
-2(-2)^2 - 3(-2) + 2 = -2(4) + 6 + 2 = -8 + 6 + 2 = 0 \quad (\text{is a solution})
\]
### Solution Set Evaluation:
After testing the values, you would identify all the values that make the equation equal to zero.
For instance:
- If testing \(x = \frac{1}{2}\) results in zero, then this is in the solution set.
- If testing \(x = -2\) results in zero, then this is also in the solution set.
Based on the results:
- If only one value results in zero, identify it and complete the statement:
> "Only \( x = \frac{1}{2} \) is in the solution set." (or however your findings conclude).
Please replace the example values provided here with the specific values you have to test against the original equation.
Let's first rewrite the equation:
\[
-2x^2 - 3x + 2 = 0
\]
We will test each proposed value in the equation. Here are the steps to check if each value is a solution:
1. Substitute the value of \(x\) into the left-hand side of the quadratic equation.
2. Evaluate the expression to see if it equals zero.
Let’s test the potential values (you'll need to provide the specific values you're checking if they are not provided):
### Example Process:
1. **Let’s assume we want to test the values.** For instance, let's say we want to test \(x = 1\):
\[
-2(1)^2 - 3(1) + 2 = -2(1) - 3 + 2 = -2 - 3 + 2 = -3 \quad (\text{not a solution})
\]
2. **Now let’s test \(x = -1\):**
\[
-2(-1)^2 - 3(-1) + 2 = -2(1) + 3 + 2 = -2 + 3 + 2 = 3 \quad (\text{not a solution})
\]
3. **Now testing \(x = \frac{1}{2}\):**
\[
-2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2 = -\frac{1}{2} - \frac{3}{2} + 2 = -2 + 2 = 0 \quad (\text{is a solution})
\]
4. **And testing \(x = 2\):**
\[
-2(2)^2 - 3(2) + 2 = -2(4) - 6 + 2 = -8 - 6 + 2 = -12 \quad (\text{not a solution})
\]
5. **Now testing \(x = -2\):**
\[
-2(-2)^2 - 3(-2) + 2 = -2(4) + 6 + 2 = -8 + 6 + 2 = 0 \quad (\text{is a solution})
\]
### Solution Set Evaluation:
After testing the values, you would identify all the values that make the equation equal to zero.
For instance:
- If testing \(x = \frac{1}{2}\) results in zero, then this is in the solution set.
- If testing \(x = -2\) results in zero, then this is also in the solution set.
Based on the results:
- If only one value results in zero, identify it and complete the statement:
> "Only \( x = \frac{1}{2} \) is in the solution set." (or however your findings conclude).
Please replace the example values provided here with the specific values you have to test against the original equation.
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