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.
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.
1 answer