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Using the Quadratic Formula to find Complex Roots
Choose Course
Algebra 2
Quadratic Functions and Equations
Using the Quadratic Formula to find Complex Roots
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Using the Quadratic Formula to find Complex Roots 1.9 - Solution
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Return to Using the Quadratic Formula to find Complex Roots
a
Let's use the
quadratic formula
to find the solutions.
x
2
−
6
x
+
1
3
=
0
UseQuadForm
Use the Quadratic Formula:
a
=
1
,
b
=
-
6
,
c
=
1
3
x
=
2
⋅
1
-
(
-
6
)
±
(
-
6
)
2
−
4
⋅
1
⋅
1
3
NegNeg
-
(
-
a
)
=
a
x
=
2
⋅
1
6
±
(
-
6
)
2
−
4
⋅
1
⋅
1
3
CalcPowProd
Calculate power and product
x
=
2
6
±
3
6
−
5
2
SubTerm
Subtract term
x
=
2
6
±
-
1
6
We have a negative number under the root sign. Therefore, the solutions will be complex.
x
=
2
6
±
-
1
6
SqrtNegToSqrtI
-
a
=
a
⋅
i
x
=
2
6
±
1
6
⋅
i
CalcRoot
Calculate root
x
=
2
6
±
4
⋅
i
SimpQuot
Simplify quotient
x
=
3
±
2
i
The solutions are
x
=
3
+
2
i
and
x
=
3
−
2
i
.
b
Now multiply the solutions together, which gives
(
3
+
2
i
)
(
3
−
2
i
)
.
This is a multiplication of two parentheses, with the only difference between them being a negative sign. That means it is a
conjugate pair of binomials
. Let's calculate its product.
(
3
+
2
i
)
(
3
−
2
i
)
ExpandDiffSquares
(
a
+
b
)
(
a
−
b
)
=
a
2
−
b
2
3
2
−
(
2
i
)
2
CalcPow
Calculate power
9
−
(
2
i
)
2
PowProdII
(
a
b
)
m
=
a
m
b
m
9
−
4
i
2
IDefPow
i
2
=
-
1
9
−
(
-
4
)
SubNeg
a
−
(
-
b
)
=
a
+
b
1
3
The product of the solutions is
1
3
.