1. Use the table to complete the statement.

Question

x Substituted Evaluate True Statement?
−34
−
3
4

−12
−
1
2

12
1
2



(1 point)
For the quadratic equation −2x2−3x+2=0
−
2

2
−
3

+
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.

Answer 1

To determine which of the given values \( x = -34, -12, 12 \) is a solution to the quadratic equation \( -2x^2 - 3x + 2 = 0 \), we will substitute each value into the equation and evaluate it.

Let's evaluate each value step by step:

1. For \( x = -34 \): \[ -2(-34)^2 - 3(-34) + 2 = -2(1156) + 102 + 2 \] \[ = -2312 + 102 + 2 \] \[ = -2312 + 104 = -2208 \quad (not : = 0) \]

2. For \( x = -12 \): \[ -2(-12)^2 - 3(-12) + 2 = -2(144) + 36 + 2 \] \[ = -288 + 36 + 2 \] \[ = -288 + 38 = -250 \quad (not : = 0) \]

3. For \( x = 12 \): \[ -2(12)^2 - 3(12) + 2 = -2(144) - 36 + 2 \] \[ = -288 - 36 + 2 \] \[ = -288 - 34 = -322 \quad (not : = 0) \]

Now that we have evaluated the function at all given values, we can conclude:

None of the substitutions yield 0, which means they are not solutions to the equation.

So the completed statement would be:

Only \( x = \text{none} \) is in the solution set.