Complex Solutions to Quadratic Equations Help
Introduction to Complex Solutions to Quadratic Equations
Every quadratic equation has a solution, real or complex. The real solutions for a quadratic equation are the x -intercepts, for the graph of the quadratic function.
The graph for f(x) = x 2 + 1 has no x -intercepts.
The equation x 2 + 1 = 0 does have two complex solutions.
Solve the equation and write the solutions in the form a + bi , where a and b are real numbers.
- 3 x 2 + 8 x + 14 = 0
In this problem, the complex solutions to the quadratic equation came in conjugate pairs. This always happens when the solutions are complex numbers. A quadratic expression that has complex zeros is called irreducible (over the reals) because it cannot be factored using real numbers. For example, the polynomial function f(x) = x 4 − 1 can be factored using real numbers as ( x 2 − 1)( x 2 + 1) = ( x − 1)( x + 1)( x 2 + 1). The factor x 2 + 1 is irreducible because it is factored as x − i )( x + i ).
We can tell which quadratic factors are irreducible without having to use the quadratic formula. We only need part of the quadratic formula, b 2 − 4 ac . When this number is negative, the quadratic factor has two complex zeros, . When this number is positive, there are two real number solutions, . When this number is zero, there is one real zero, . For this reason, b 2 − 4 ac is called the discriminant .
The graphs of some polynomials having irreducible quadratic factors need extra points plotted to get a more accurate graph. The graph in Figure 7.17 shows the graph of f(x) = x 4 − 3 x 2 − 4 using our usual method—plotting the x -intercepts, a point to the left of the smallest x -intercept, a point between each consecutive pair of x -intercepts, and a point to the right of the largest x -intercept.
See what happens to the graph when we plot the points for x = 1 and x = − 1.
The graph of f(x) = ( x − 2)( x 2 + 6 x + 10) is sketched in Figure 7.19. The graphs we have sketched have several vertices between x -intercepts. When this happens, we need calculus to find them.
Practice problems for this concept can be found at: Polynomial Functions Practice Test.
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