For the quadratic equation - 2x2 - 32 + 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.
To complete the table, we are given the quadratic equation -2x^2 - 32x + 2 = 0. We need to test each of the given values to see if they satisfy this equation.
For example, let's test x = -3:
-2(-3)^2 - 32(-3) + 2 = -18 + 96 + 2 = 80.
Since the equation does not equal zero when x = -3, this value is not a solution.
Similarly, let's test x = 0:
-2(0)^2 - 32(0) + 2 = 2.
Again, since the equation does not equal zero when x = 0, this value is also not a solution.
Lastly, let's test x = 2:
-2(2)^2 - 32(2) + 2 = -8 - 64 + 2 = -70.
This time, since the equation equals zero when x = 2, this value is a solution.
Therefore, the value x = 2 is in the solution set of the quadratic equation.
To complete the table and test each value, we need to substitute each value into the quadratic equation and check if the equation equals zero. Let's start by substituting the given values into the quadratic equation: -2x^2 - 32 + 2 = 0
First, let's test x = 0:
By substituting x = 0 into the equation, we get:
-2(0)^2 - 32 + 2 = 0
0 - 32 + 2 = 0
-32 + 2 = 0
-30 = 0
Since -30 does not equal 0, x = 0 is not a solution.
Now, let's test x = 1:
By substituting x = 1 into the equation, we get:
-2(1)^2 - 32 + 2 = 0
-2(1) - 32 + 2 = 0
-2 - 32 + 2 = 0
-34 + 2 = 0
-32 = 0
Since -32 does not equal 0, x = 1 is not a solution.
Lastly, let's test x = -4:
By substituting x = -4 into the equation, we get:
-2(-4)^2 - 32 + 2 = 0
-2(16) - 32 + 2 = 0
-32 - 32 + 2 = 0
-64 + 2 = 0
-62 = 0
Since -62 does not equal 0, x = -4 is not a solution.
None of the given values x = 0, x = 1, or x = -4 are solutions to the quadratic equation -2x^2 - 32 + 2 = 0.
To complete the table for the given quadratic equation, we need to substitute each value into the equation and determine whether it satisfies the equation (i.e., if it results in the equation being true).
Given quadratic equation: -2x^2 - 32x + 2 = 0
Let's substitute the given values into the equation one by one and check whether they are solutions:
Value | Substituting into the equation | Result
-----------------------------------------------------
0 |-2(0)^2 - 32(0) + 2 = 0 | 2 ≠ 0 (False)
1 |-2(1)^2 - 32(1) + 2 = 0 | -34 ≠ 0 (False)
-1 |-2(-1)^2 - 32(-1) + 2 = 0 | 30 ≠ 0 (False)
2 |-2(2)^2 - 32(2) + 2 = 0 | -82 ≠ 0 (False)
None of the given values satisfy the equation. Therefore, there is no solution that can be found from the given values for this quadratic equation.