Algebra II: Solving Quadratic Equation with Imaginary Roots
This is a free lesson from our course in Algebra II
 
   
This lesson explains how to solve quadratic equation that has imaginary roots. When you look at the write up; may find difficult to comprehend but once go through some examples with solution, and the instructor's explanation with the help of audio, video presentation in own hand writing, you will find it easier.
A root of an equation is a solution of that equation. When the roots of a quadratic equation are imaginary, they always occur in conjugate pairs i.e. the two solutions of equation are complex conjugates of each other.You have learnt earlier that the discriminant of the quadratic formula is
      b2 - 4ac.
If the discriminant is negative, when you solve the quadratic equation it forms complex roots. (More text below video...)
<h2> Solving quadratic equation with imaginary roots - Watch video (Algebra II)</h2> <p> linear, equations, video, variable, example, Algebra II, substitution, system of equation, solving equation, equations, coefficients, math help, equations in three variables, practice questions, quizzes</p> <p> The method of solving system of equations having imaginary roots.</p>
Other useful lessons:
Imaginary Unit
Plotting complex numbers on a 2D plane
Adding and Subtracting Complex Numbers
Multiply and Divide the complex numbers
Classification of roots using the discriminant
(Continued from above) You can find out the roots using the quadratic formula applying
      x = {-b (b2 4ac )}/ 2a.
Since discriminant is a negative number, if the given equation is to be solved there would be two conjugate complex (imaginary) solutions. To write the answer in a + bi form remove the common factors and then write b as a coefficient of i. For example, find the roots of the quadratic equation 3x2 + 4x + 8 = 0.
      x = {-4 (42 4(3)(8) )}/ 2(3).
      x = {-4 (16 96 )}/ 6
      x = {-4 (80)}/ 6
      x = {-4 4i5}/6
      x = {-2 2i5}/ 3
      x = -2/3 + (2i5)/3, -2/3 - (2i5)/3.
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