Advanced Algebra with Trig
Lesson A.6b Notes
Name: ____________________________________
Solving Quadratic Equations with a Negative Discriminant
We have discussed how a quadratic equation with a negative discriminant has no real number solution. For
example,
Cannot square root a negative in
Ex 1: x 2  4  0
x 2  4
x   4
the real number system
However,
if we extend our
number system to allow complex numbers, quadratic equations will always have a
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solution. Thus, if N is a positive real number, we define the principal square root of -N, denoted N , as
N 
Ex 2.
Ex 3.
Ex 4.
1  1i  i
4  4i  2i
8  8i  2 2i
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 Practice 1:
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Practice 2:
Ni
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3 
18 
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Let’s apply this to solving quadratic equations in the complex number system.
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Ex 5. Solve x 2  4  0
x2  4  0
x 2  4
x   4
x  2i
x   4i
Practice
3: Solve x 2  9  0
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Now let’s look 
at quadratic equations where we must complete the square to solve.
Ex 6. Solve x 2  4 x  8  0 in the complex number system
x 2  4 x  4  8  4
(x  2) 2  4
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x  2   4
x  2  2i
x  2  2i
Practice 4: Solve x 2  2x  4  0 in the complex number system
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Advanced Algebra with Trig
Lesson A.6b Notes
Character of the Solutions of a Quadratic Equation
In the complex number system, consider a quadratic equation ax 2  bx  c  0 with real coefficients.
1. If b 2  4ac  0 , the equation has two unequal real solutions
2. If b 2  4ac  0 , the equation has a repeated
real solution, a double root

2
3. If b  4ac  0 , the equation has two complex (conjugate) solutions that are not real.
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Ex 7. Given x 2  4 x  8 from example 6, we can determine the character of its solution(s)
b 2  4ac  (4)2  4(1)(8)  16  0
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Thus, two complex solutions
This matches our solution of x  2  2i from example 6.
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Ex 8. Determine the character of the solutions for 9x 2  6x  1  0 , then solve for x.
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Thus, repeated real root
b 2  4ac  (6)2  4(9)(1)  0
We can factor to solve the equation
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9x 2  6x  1  0
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(3x 1)(3x 1)  0
x  13 , x  13
Repeated real root!
Ex 9. Determine the character of the solutions for x 2  4 x  1  0 , then solve for x.
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Thus, two unequal real solutions
b2  4ac  (4)2  4(1)(1) 12  0
We can complete the square to solve
 the equation
x 2  4x  4  1 4
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(x  2)2  3
x 2  3
x  2 3
Two unequal real solutions!
Practice 5: Determine the character of the solutions for x 2  6x  13  0, then solve for x.
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