Use the table to answer the question.

Question

x Substituted Evaluate True Statement?
−34
−12
12
​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.​

(1 point)
Only x=blank is in the solution set.

Answer 1

To determine which values are solutions for the quadratic equation \(-2x^2 - 3x + 2 = 0\), you need to substitute each value of \(x\) into the equation and check whether the equation holds true (i.e., whether the left-hand side equals zero).

Substituting each value:

1. For \(x = -34\): \[ -2(-34)^2 - 3(-34) + 2 = -2(1156) + 102 + 2 = -2312 + 102 + 2 = -2210 \quad (\text{Not } 0) \]

2. For \(x = -12\): \[ -2(-12)^2 - 3(-12) + 2 = -2(144) + 36 + 2 = -288 + 36 + 2 = -250 \quad (\text{Not } 0) \]

3. For \(x = 12\): \[ -2(12)^2 - 3(12) + 2 = -2(144) - 36 + 2 = -288 - 36 + 2 = -322 \quad (\text{Not } 0) \]

Summary of Evaluations:

• For \(x = -34\): Result = -2210 (Not a solution)
• For \(x = -12\): Result = -250 (Not a solution)
• For \(x = 12\): Result = -322 (Not a solution)

Conclusion:

None of the values provided \(x = -34\), \(x = -12\), or \(x = 12\) are valid solutions to the equation \(-2x^2 - 3x + 2 = 0\). Therefore,

Only x=blank is in the solution set.

Since none of the provided values solve the equation, we would fill in "none" or leave it blank, as none are in the solution set.

Answer 2

you have to choose one

Answer 3

I apologize for the oversight. Let's solve the equation correctly using the quadratic formula to find potential solutions.

The quadratic equation is:

\[ -2x^2 - 3x + 2 = 0. \]

To use the quadratic formula, we first rewrite the equation in standard form \(ax^2 + bx + c = 0\):

\[ 2x^2 + 3x - 2 = 0. \]

Where:

• \(a = 2\)
• \(b = 3\)
• \(c = -2\)

The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]

Step 1: Calculate the Discriminant

\[ b^2 - 4ac = 3^2 - 4(2)(-2) = 9 + 16 = 25. \]

Step 2: Apply the Quadratic Formula

\[ x = \frac{-3 \pm \sqrt{25}}{2 \cdot 2} = \frac{-3 \pm 5}{4}. \]

Step 3: Calculate the Two Solutions

1. First Solution:

\[ x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}. \]

2. Second Solution:

\[ x = \frac{-3 - 5}{4} = \frac{-8}{4} = -2. \]

Conclusion

The solutions to the equation \(-2x^2 - 3x + 2 = 0\) are \(x = \frac{1}{2}\) and \(x = -2\).

Comparison with Given Values

• Given Values: \(-34\), \(-12\), \(12\)

Since none of those values are \(\frac{1}{2}\) or \(-2\), we can now pick one of the provided values for which the left-hand side evaluates to the least in the interest of this query.

Thus, we can conclude:

Only x=\(-12\) (the least but not a solution) is in the solution set.

However, please note that none of the stated values are valid solutions for the given quadratic equation. If you are strictly required to choose among given values, then x = -12 will suffice despite not being a solution.

Answer 4

Use the table to answer the question.

x Substituted Evaluate True Statement?
−3/4
−1/2
1/2
​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.​

(1 point)
Only x=blank is in the solution set.

Answer 5

Let's substitute each of the given values for \(x\) into the quadratic equation \(-2x^2 - 3x + 2 = 0\) and determine which, if any, satisfy the equation.

Given Values:

1. \(x = -\frac{3}{4}\)
2. \(x = -\frac{1}{2}\)
3. \(x = \frac{1}{2}\)

Evaluating Each Value

1. For \(x = -\frac{3}{4}\): \[ -2\left(-\frac{3}{4}\right)^2 - 3\left(-\frac{3}{4}\right) + 2 = -2\left(\frac{9}{16}\right) + \frac{9}{4} + 2 = -\frac{18}{16} + \frac{9 \cdot 4}{16} + \frac{2 \cdot 16}{16} = -\frac{18}{16} + \frac{36}{16} + \frac{32}{16} = \frac{-18 + 36 + 32}{16} = \frac{50}{16} = \frac{25}{8} \quad (\text{Not } 0) \]

2. For \(x = -\frac{1}{2}\): \[ -2\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) + \frac{3}{2} + 2 = -\frac{2}{4} + \frac{3}{2} + 2 = -\frac{1}{2} + \frac{3}{2} + 2 = \frac{-1 + 3 + 4}{2} = \frac{6}{2} = 3 \quad (\text{Not } 0) \]

3. For \(x = \frac{1}{2}\): \[ -2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2 = -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2 = -\frac{2}{4} - \frac{3}{2} + 2 = -\frac{1}{2} - \frac{3}{2} + 2 = \frac{-1 - 3 + 4}{2} = \frac{0}{2} = 0 \quad (\text{True Statement}) \]

Conclusion from Evaluations

• For \(x = -\frac{3}{4}\): Result = \(\frac{25}{8}\) (Not a solution)
• For \(x = -\frac{1}{2}\): Result = \(3\) (Not a solution)
• For \(x = \frac{1}{2}\): Result = \(0\) (This is a solution)

Final Answer

Only \(x = \frac{1}{2}\) is in the solution set.

Answer 6

Assuming an equation with one side as a squared variable expression and the other side as a numeric expression, which of the following statements is correct?

Statement #1: If the numeric expression is zero, there are two solutions.

Statement #2: If the numeric expression is zero, there is one solution.

Statement #3: If the numeric expression is zero, there are no solutions.

Answer 7

Let's analyze the situation when you have an equation of the form:

\[ x^2 = k \]

where \(k\) is a numeric expression. We can consider the cases where \(k = 0\) and analyze the three statements based on this condition.

Case: \(k = 0\)

The equations would be:

\[ x^2 = 0 \]

The solution to this equation is:

\[ x = 0 \]

Evaluation of Statements:

1. Statement #1: If the numeric expression is zero, there are two solutions.

   • Incorrect. There is only one solution when \(x^2 = 0\).

2. Statement #2: If the numeric expression is zero, there is one solution.

   • Correct. As determined, \(x^2 = 0\) has exactly one solution: \(x = 0\).

3. Statement #3: If the numeric expression is zero, there are no solutions.

   • Incorrect. There is one solution, \(x = 0\).

Conclusion

The correct statement is:

Statement #2: If the numeric expression is zero, there is one solution.

Answer 8

Solve x2=78 by inspection. There are two real solutions. Enter the lesser number first, and use exponents in your responses. (1 point)
{blank,blank}

Answer 9

To solve the equation \(x^2 = 78\) by inspection, we need to find the values of \(x\) such that their squares equal 78.

Since we know that \(x^2 = 78\), we take the square root of both sides:

\[ x = \pm \sqrt{78}. \]

Now, let's simplify \(\sqrt{78}\):

\[ \sqrt{78} = \sqrt{39 \times 2} = \sqrt{39} \sqrt{2} \approx 8.83. \]

Thus, the two solutions are:

1. \(x = -\sqrt{78}\) (the lesser number)
2. \(x = \sqrt{78}\)

The solutions in order with exponents would be:

\[ {-\sqrt{78}, \sqrt{78}}. \]

So, the final answer is:

\[ {-\sqrt{78}, \sqrt{78}}. \]

Answer 10

Solve the following quadratic equation using square roots: −9+(x−3)2−4=−4 .(1 point)
Responses

{28,−22}
left brace 28 comma negative 22 right brace

{2,−8}
left brace 2 comma negative 8 right brace

{16,−10}
left brace 16 comma negative 10 right brace

{8,−2}

Answer 11

To solve the quadratic equation \(-9 + (x - 3)^2 - 4 = -4\), we will first simplify and isolate the squared term.

Step 1: Simplify the equation

Starting from the equation:

\[ -9 + (x - 3)^2 - 4 = -4 \]

Add 9 to both sides to start simplifying:

\[ (x - 3)^2 - 4 = 5 \]

Next, add 4 to both sides:

\[ (x - 3)^2 = 9 \]

Step 2: Solve by taking the square root

Now we take the square root of both sides:

\[ x - 3 = \pm 3 \]

Step 3: Solve for \(x\)

This gives us two equations to solve:

1. For the positive case:

\[ x - 3 = 3 \quad \Rightarrow \quad x = 6 \]

2. For the negative case:

\[ x - 3 = -3 \quad \Rightarrow \quad x = 0 \]

Step 4: Final solution set

The solutions are:

\[ {6, 0} \]

None of the listed response options match \({6, 0}\). Let's check if any solution set provided corresponds to the numbers we arrived at.

However, considering the instructions, we might want to verify our result against the options provided:

• {28, -22}
• {2, -8}
• {16, -10}
• {8, -2}

None of these pairs appear to match. Therefore, can we conclude that the question might have either incorrect response options or an error in processing? The calculated solutions are accurate given the original equation.

If you believe adjustments could be necessary in either the problem statement or in response options, please double-check the source. As per the current work:

The correct solutions are \(x = 6\) and \(x = 0\) which corresponds to {6, 0}.