Detailed solutions and explanations to the questions in Solve Quadratic Equations Using Discriminants.
Given the quadratic equation
\[ x^2 - 3x + 2 = 0 \]Identify the coefficients:
\[ a = 1, \quad b = -3, \quad c = 2 \]Discriminant:
\[ D = b^2 - 4ac = (-3)^2 - 4(1)(2) = 1 \]Since \(D > 0\), there are two real solutions:
\[ x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + 1}{2} = 2 \] \[ x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - 1}{2} = 1 \]Given
\[ \frac{x^2}{2} = -8 - 4x \]Multiply both sides by 2 and simplify:
\[ x^2 + 8x + 16 = 0 \]Coefficients:
\[ a = 1, \quad b = 8, \quad c = 16 \] \[ D = b^2 - 4ac = 8^2 - 4(1)(16) = 0 \]Since \(D = 0\), there is one real solution:
\[ x = \frac{-b}{2a} = \frac{-8}{2} = -4 \]Given
\[ x^2 - 4x + 5 = 0 \]Coefficients:
\[ a = 1, \quad b = -4, \quad c = 5 \] \[ D = b^2 - 4ac = (-4)^2 - 4(1)(5) = -4 \]Since \(D < 0\), there are two complex solutions:
\[ x = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{-4}}{2} = 2 + i \] \[ x = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{-4}}{2} = 2 - i \]where \(i = \sqrt{-1}\) is the imaginary unit.
Given
\[ x^2 + x + m + 1 = 0 \]Coefficients:
\[ a = 1, \quad b = 1, \quad c = m+1 \] \[ D = b^2 - 4ac = 1 - 4(m+1) = -3 - 4m \]For one solution, \(D = 0\):
\[ -3 - 4m = 0 \quad \Rightarrow \quad m = -\frac{3}{4} \]For two real solutions, \(D > 0\):
\[ -3 - 4m > 0 \quad \Rightarrow \quad m \in (-\infty, -3/4) \]For two complex solutions, \(D < 0\):
\[ -3 - 4m < 0 \quad \Rightarrow \quad m \in (-3/4, +\infty) \]