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.
To determine which of the given values is a solution to the quadratic equation \( -2x^2 - 3x + 2 = 0 \), we can substitute each value for \( x \) into the equation and check if the equation is satisfied (equal to zero).
1. For \( x = -34 \):
\( -2(-34)^2 - 3(-34) + 2 = -2(1156) + 102 + 2 = -2312 + 102 + 2 = -2312 + 104 = -2208 \)
This is not equal to zero, so \( x = -34 \) is not a solution.
2. For \( x = -12 \):
\( -2(-12)^2 - 3(-12) + 2 = -2(144) + 36 + 2 = -288 + 36 + 2 = -288 + 38 = -250 \)
This is not equal to zero, so \( x = -12 \) is not a solution.
3. For \( x = 12 \):
\( -2(12)^2 - 3(12) + 2 = -2(144) - 36 + 2 = -288 - 36 + 2 = -288 - 34 = -322 \)
This is not equal to zero, so \( x = 12 \) is not a solution.
Since none of the three values provided are solutions to the equation \( -2x^2 - 3x + 2 = 0 \), we can confirm that none of the given options are in the solution set.
We can actually find the correct solutions by factoring the quadratic equation if possible or by using the quadratic formula.
To factor the quadratic equation:
\( -2x^2 - 3x + 2 = 0 \)
Factor the quadratic expression:
\( (-2x - 1)(x - 2) = 0 \)
Setting each factor equal to zero gives us the solutions for \( x \):
\( -2x - 1 = 0 \) or \( x - 2 = 0 \)
\( -2x = 1 \) or \( x = 2 \)
\( x = -\frac{1}{2} \) or \( x = 2 \)
So, the solutions to the equation are \( x = -\frac{1}{2} \) and \( x = 2 \), and neither of these is in the set \{ -34, -12, 12 \}.