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.
Question 2
2. Solve x2=289
by inspection. There are two real solutions. Enter the lesser number first.
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{
,
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, 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.
Question 2
2. Solve x2=289
by inspection. There are two real solutions. Enter the lesser number first.
Desmos Calculator (click here)
(2 points)
{
,
}
Answers
GPT-4o mini
Let's start with the first part of your question regarding the quadratic equation \(-2x^2 - 3x + 2 = 0\).
To determine if a value is a solution of this equation, we can substitute the given values of \(x\) into the equation to see if the left side equals zero.
### Step 1: Solve for solutions using the quadratic formula
First, we can find the exact solutions to the quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case:
- \(a = -2\)
- \(b = -3\)
- \(c = 2\)
Calculating \(b^2 - 4ac\):
\[
b^2 - 4ac = (-3)^2 - 4(-2)(2) = 9 + 16 = 25
\]
Now, substituting into the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{25}}{2(-2)} = \frac{3 \pm 5}{-4}
\]
Calculating the two possible solutions:
1. \(x = \frac{3 + 5}{-4} = \frac{8}{-4} = -2\)
2. \(x = \frac{3 - 5}{-4} = \frac{-2}{-4} = \frac{1}{2}\)
Thus, the solutions to the equation are \(x = -2\) and \(x = \frac{1}{2}\).
### Step 2: Testing the Values
To complete the table, let's test some specific values (let's say you have \(x = -2\) and \(x = \frac{1}{2}\) and additional values, e.g., \(0, 1, -3\), or any others). You would substitute each of those into \(-2x^2 - 3x + 2\) and evaluate.
For example:
1. **For \(x = -2\)**:
\[
-2(-2)^2 - 3(-2) + 2 = -2(4) + 6 + 2 = -8 + 6 + 2 = 0
\]
(Solution)
2. **For \(x = \frac{1}{2}\)**:
\[
-2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2 = -\frac{1}{2} - \frac{3}{2} + 2 = -2 + 2 = 0
\]
(Solution)
3. **For \(x = 0\)**:
\[
-2(0)^2 - 3(0) + 2 = 0 + 0 + 2 = 2
\]
(Not a solution)
4. **For \(x = 1\)**:
\[
-2(1)^2 - 3(1) + 2 = -2 - 3 + 2 = -3
\]
(Not a solution)
5. **For \(x = -3\)**:
\[
-2(-3)^2 - 3(-3) + 2 = -2(9) + 9 + 2 = -18 + 9 + 2 = -7
\]
(Not a solution)
Given the evaluations above, the values \(x = -2\) and \(x = \frac{1}{2}\) are indeed in the solution set.
### Answer for Question 1:
Only \(x = -2\) and \(x = \frac{1}{2}\) are in the solution set.
---
### Question 2: Solve \(x^2 = 289\) by inspection
To solve \(x^2 = 289\), we can take the square root of both sides.
\[
x = \pm \sqrt{289}
\]
Calculating \(\sqrt{289}\):
\[
\sqrt{289} = 17
\]
Thus, the two solutions are:
\[
x = 17 \quad \text{and} \quad x = -17
\]
### Final Answer for Question 2:
\[
\{ -17, 17 \}
\]
---
So your complete answers would be:
**Question 1:** Only \(x = -2\) and \(x = \frac{1}{2}\) are in the solution set.
**Question 2:** \(\{-17, 17\}\).
To determine if a value is a solution of this equation, we can substitute the given values of \(x\) into the equation to see if the left side equals zero.
### Step 1: Solve for solutions using the quadratic formula
First, we can find the exact solutions to the quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case:
- \(a = -2\)
- \(b = -3\)
- \(c = 2\)
Calculating \(b^2 - 4ac\):
\[
b^2 - 4ac = (-3)^2 - 4(-2)(2) = 9 + 16 = 25
\]
Now, substituting into the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{25}}{2(-2)} = \frac{3 \pm 5}{-4}
\]
Calculating the two possible solutions:
1. \(x = \frac{3 + 5}{-4} = \frac{8}{-4} = -2\)
2. \(x = \frac{3 - 5}{-4} = \frac{-2}{-4} = \frac{1}{2}\)
Thus, the solutions to the equation are \(x = -2\) and \(x = \frac{1}{2}\).
### Step 2: Testing the Values
To complete the table, let's test some specific values (let's say you have \(x = -2\) and \(x = \frac{1}{2}\) and additional values, e.g., \(0, 1, -3\), or any others). You would substitute each of those into \(-2x^2 - 3x + 2\) and evaluate.
For example:
1. **For \(x = -2\)**:
\[
-2(-2)^2 - 3(-2) + 2 = -2(4) + 6 + 2 = -8 + 6 + 2 = 0
\]
(Solution)
2. **For \(x = \frac{1}{2}\)**:
\[
-2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2 = -\frac{1}{2} - \frac{3}{2} + 2 = -2 + 2 = 0
\]
(Solution)
3. **For \(x = 0\)**:
\[
-2(0)^2 - 3(0) + 2 = 0 + 0 + 2 = 2
\]
(Not a solution)
4. **For \(x = 1\)**:
\[
-2(1)^2 - 3(1) + 2 = -2 - 3 + 2 = -3
\]
(Not a solution)
5. **For \(x = -3\)**:
\[
-2(-3)^2 - 3(-3) + 2 = -2(9) + 9 + 2 = -18 + 9 + 2 = -7
\]
(Not a solution)
Given the evaluations above, the values \(x = -2\) and \(x = \frac{1}{2}\) are indeed in the solution set.
### Answer for Question 1:
Only \(x = -2\) and \(x = \frac{1}{2}\) are in the solution set.
---
### Question 2: Solve \(x^2 = 289\) by inspection
To solve \(x^2 = 289\), we can take the square root of both sides.
\[
x = \pm \sqrt{289}
\]
Calculating \(\sqrt{289}\):
\[
\sqrt{289} = 17
\]
Thus, the two solutions are:
\[
x = 17 \quad \text{and} \quad x = -17
\]
### Final Answer for Question 2:
\[
\{ -17, 17 \}
\]
---
So your complete answers would be:
**Question 1:** Only \(x = -2\) and \(x = \frac{1}{2}\) are in the solution set.
**Question 2:** \(\{-17, 17\}\).