This tutorial presents questions and comprehensive explanations on solving quadratic equations using the discriminant and the quadratic formula. Each solution provides detailed step-by-step workings to help students understand how the sign of minibars/discriminants determines the number and type of solutions (real or complex roots).
Topics covered in this guide include:
A quadratic equation in one variable is written in standard form as:
\[ ax^2 + bx + c = 0 \] where \(a \neq 0\) and \(a, b, c\) are real constants.
The solutions can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The expression under the square root is called the discriminant (\(\Delta\)):
\[ \Delta = b^2 - 4ac \]
The discriminant determines the number and nature of the solutions:
Question 1: Solve the quadratic equation \(x^2 + 3x = 4\).
Rewrite with zero on the right-hand side:
\[ x^2 + 3x - 4 = 0 \]Identify coefficients: \(a = 1, \; b = 3, \; c = -4\).
Calculate the discriminant:
\[ \Delta = b^2 - 4ac = 3^2 - 4(1)(-4) = 9 + 16 = 25 \]Since \(\Delta > 0\), there are two distinct real solutions:
\[ x_1 = \frac{-3 + \sqrt{25}}{2(1)} = \frac{-3 + 5}{2} = 1, \quad x_2 = \frac{-3 - \sqrt{25}}{2(1)} = \frac{-3 - 5}{2} = -4 \]Check: For \(x = 1 \Rightarrow 1^2 + 3(1) = 4\). For \(x = -4 \Rightarrow (-4)^2 + 3(-4) = 4\).
Conclusion: Solutions are \(x = 1, -4\).
Question 2: Solve the equation \(\dfrac{x^2}{3} + 3 = 2x\).
Multiply through by 3 to clear the fraction:
\[ x^2 + 9 = 6x \implies x^2 - 6x + 9 = 0 \]Identify coefficients: \(a = 1, \; b = -6, \; c = 9\).
Discriminant:
\[ \Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0 \]Since \(\Delta = 0\), there is one real solution:
\[ x = \frac{-b}{2a} = \frac{6}{2(1)} = 3 \]Check: \(\dfrac{3^2}{3} + 3 = 3 + 3 = 6\), and \(2(3) = 6\).
Conclusion: One real solution: \(x = 3\).
Question 3: Solve the equation \(x^2 - 4x + 13 = 0\).
Identify coefficients: \(a = 1, \; b = -4, \; c = 13\).
Discriminant:
\[ \Delta = (-4)^2 - 4(1)(13) = 16 - 52 = -36 \]Since \(\Delta < 0\), the solutions are complex conjugates:
\[ x_1 = \frac{4 + \sqrt{-36}}{2} = \frac{4 + 6i}{2} = 2 + 3i \] \[ x_2 = \frac{4 - \sqrt{-36}}{2} = \frac{4 - 6i}{2} = 2 - 3i \]Conclusion: Two complex conjugate solutions: \(x = 2 + 3i\) and \(x = 2 - 3i\).
Question 4: Determine the number and type of solutions for the parameter equation \(x^2 + mx + 1 = 0\) for different values of \(m\).
Identify coefficients: \(a = 1, \; b = m, \; c = 1\).
Discriminant: \(\Delta = m^2 - 4\).
Question 5: Solve the quadratic equation \(x^2 - 3x + 2 = 0\).
Identify coefficients: \(a = 1, \; b = -3, \; c = 2\).
Discriminant:
\[ \Delta = (-3)^2 - 4(1)(2) = 9 - 8 = 1 \]Since \(\Delta > 0\), two real solutions exist:
\[ x_1 = \frac{3 + \sqrt{1}}{2} = 2, \quad x_2 = \frac{3 - \sqrt{1}}{2} = 1 \]Conclusion: Solutions: \(x = 1, 2\).
Question 6: Solve the equation \(\dfrac{x^2}{2} = -8 - 4x\).
Multiply both sides by 2 and rearrange into standard form:
\[ x^2 = -16 - 8x \implies x^2 + 8x + 16 = 0 \]Identify coefficients: \(a = 1, \; b = 8, \; c = 16\).
Discriminant:
\[ \Delta = 8^2 - 4(1)(16) = 64 - 64 = 0 \]Since \(\Delta = 0\), there is one real solution:
\[ x = \frac{-8}{2(1)} = -4 \]Conclusion: One real solution: \(x = -4\).
Question 7: Solve the equation \(x^2 - 4x + 5 = 0\).
Identify coefficients: \(a = 1, \; b = -4, \; c = 5\).
Discriminant:
\[ \Delta = (-4)^2 - 4(1)(5) = 16 - 20 = -4 \]Since \(\Delta < 0\), there are two complex solutions:
\[ x = \frac{4 \pm \sqrt{-4}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i \]where \(i = \sqrt{-1}\) is the imaginary unit.
Question 8: Find the values of parameter \(m\) for which the equation \(x^2 + x + m + 1 = 0\) has:
Identify coefficients: \(a = 1, \; b = 1, \; c = m + 1\).
Discriminant:
\[ \Delta = 1^2 - 4(1)(m + 1) = 1 - 4m - 4 = -3 - 4m \]