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Matched Exercise 1: Find all values of the parameter m in the quadratic equation
x 2 + x + m + 1 = 0
such that the equation has
- one solution,
- 2 real solutions, and
- 2 complex solutions.
Solution to Matched Exercise 1:
- Given
x 2 + x + m + 1 = 0
- Find the discriminant>
D = b2 - 4ac = 12 - 4(1)(m + 1) = -3 - 4m
- For the equation to have one solution, the discriminant has to be equal to zero.
-3 - 4m = 0
- Solve the above equation for m.
m = -3/4
- For the equation to have 2 real solution, the discriminant has to be greater than zero.
-3 - 4m > 0
- The inequality -3 - 4m > 0 has the following solution set.
(-infinity , -3/4)
- For the equation to have 2 complex solution, the discriminant has to be less than zero.
-3 - 4m < 0
- The inequality -3 - 4m > 0 has the following solution set.
(-3/4, +infinity)
More references and links on how to Solve Equations, Systems of Equations and Inequalities.
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