H Algebra 2/Trig · Unit 2
Use D = b² − 4ac to name the roots, or to find k so there is one real root or two imaginary roots.
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The discriminant
x = −b ± √(b² − 4ac)2a
D = b² − 4ac
One real root
When D = 0, there is one real root. The graph touches the x-axis at one point.
Example: x² − 4x + 4 = 0
The formula is
D = b² − 4ac
Substitute.
D = (−4)² − 4(1)(4)
D = 16 − 16
D = 0
One real root.
Two real roots
When D > 0, there are two real roots. The graph crosses the x-axis at two points.
Example: x² − 3x + 2 = 0
The formula is
D = b² − 4ac
Substitute.
D = (−3)² − 4(1)(2)
D = 9 − 8
D = 1
Two real roots.
No real roots
When D < 0, there are two imaginary roots (0 real roots). The graph does not cross the x-axis.
Example: x² + 2x + 2 = 0
The formula is
D = b² − 4ac
Substitute.
D = (2)² − 4(1)(2)
D = 4 − 8
D = −4
Two imaginary roots (0 real roots).
Compute D and name the roots.
The formula is
D = b² − 4ac
Substitute.
D = (−3)² − 4(2)(1)
D = 9 − 8
D = 1
Two real roots.
Answer: D = 1Two real roots.
The formula is
D = b² − 4ac
Substitute.
D = (1)² − 4(1)(4)
D = 1 − 16
D = −15
Two imaginary roots (0 real roots).
Answer: D = −15Two imaginary roots (0 real roots).
The formula is
D = b² − 4ac
Substitute.
D = (−8)² − 4(1)(16)
D = 64 − 64
D = 0
One real root.
Answer: D = 0One real root.
The formula is
D = b² − 4ac
Substitute.
D = (−5)² − 4(1)(6)
D = 25 − 24
D = 1
Two real roots.
Answer: D = 1Two real roots.
The formula is
D = b² − 4ac
Substitute.
D = (2)² − 4(3)(5)
D = 4 − 60
D = −56
Two imaginary roots (0 real roots).
Answer: D = −56Two imaginary roots (0 real roots).
The formula is
D = b² − 4ac
Substitute.
D = (6)² − 4(1)(9)
D = 36 − 36
D = 0
One real root.
Answer: D = 0One real root.
The equation is not set equal to zero. Rearrange, then find D.
Move every term to one side.
x² − 2x + 3 = 0
The formula is
D = b² − 4ac
Substitute.
D = (−2)² − 4(1)(3)
D = 4 − 12
D = −8
Two imaginary roots (0 real roots).
Answer: D = −8Two imaginary roots (0 real roots).
Move every term to one side.
4x² − 12x + 9 = 0
The formula is
D = b² − 4ac
Substitute.
D = (−12)² − 4(4)(9)
D = 144 − 144
D = 0
One real root.
Answer: D = 0One real root.
Move every term to one side.
2x² − 5x − 3 = 0
The formula is
D = b² − 4ac
Substitute.
D = (−5)² − 4(2)(−3)
D = 25 − (−24)
D = 25 + 24
D = 49
Two real roots.
Answer: D = 49Two real roots.
Move every term to one side.
x² − 4x + 7 = 0
The formula is
D = b² − 4ac
Substitute.
D = (−4)² − 4(1)(7)
D = 16 − 28
D = −12
Two imaginary roots (0 real roots).
Answer: D = −12Two imaginary roots (0 real roots).
Move every term to one side.
3x² + x − 4 = 0
The formula is
D = b² − 4ac
Substitute.
D = (1)² − 4(3)(−4)
D = 1 − (−48)
D = 1 + 48
D = 49
Two real roots.
Answer: D = 49Two real roots.
Move every term to one side.
x² + 8x + 16 = 0
The formula is
D = b² − 4ac
Substitute.
D = (8)² − 4(1)(16)
D = 64 − 64
D = 0
One real root.
Answer: D = 0One real root.
Find k so the quadratic has one real root, two real roots, or two imaginary roots.
One real root means D = 0.
The formula is
D = b² − 4ac
a = k, b = 10, c = −8
100 − 4(k)(−8) = 0
100 + 32k = 0
32k = −100
k = −258
Answer: k = −258
Two imaginary roots (0 real roots) mean D < 0, and k ≠ 0.
D = 36 − 36k
36 − 36k < 0
−36k < −36
k > 1
Answer: k > 1
D = 0.
k² − 64 = 0
k² = 64
k = 8 or k = −8
Answer: k = 8 or k = −8
D = 0.
k² − 4(2)(5) = 0
k² − 40 = 0
k² = 40
k = ±2√10
Answer: k = ±2√10
Two real roots mean D > 0.
16 − 12k > 0
−12k > −16
k < 43
Answer: k < 43
D < 0, and k cannot be 0.
64 − 8k < 0
−8k < −64
k > 8
Answer: k > 8
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