SME
Grade 11 Math · Problem type
Complex solutions of a quadratic
Use this for “Solve x² − 4x + 13 = 0.”
Diagram
The conjugate z̄ = 2 − 3i: z reflected across the real axis.
This is an example. Type your own values to start a new problem.
a
x²
+
(
b
)
x
+
c
=
0
D
Discriminant b² − 4ac
worked out (not typed)
p
Real part
worked out (not typed)
q
Imaginary part
worked out (not typed)
Formulas
D
=
b
2
−
4
×
a
×
c
−36
=
(−4)
2
−
4
×
1
×
13
p
=
−
b
2 ×
a
2
=
−
−4
2 × 1
q
=
√
−1 ×
D
2 × |
a
|
3
=
√
−1 × (−36)
2 × |1|
Assumptions
•
When b² − 4ac is negative, the square root in the quadratic formula is imaginary: √(−36) = 6i.
•
The two solutions are p + qi and p − qi, a conjugate pair.
•
Check by putting p + qi back into the equation, with i² = −1.
Step-by-step
WE KNOW
a = 1, b = −4, c = 13
FIND
discriminant b² − 4ac (D), real part (p), imaginary part (q)
Step 1 · Find discriminant b² − 4ac (D)
Use D = b² − 4 × a × c
The discriminant: negative means no real solutions.
D
=
(−4)
2
−
4
×
1
×
13
D
=
16
−
4
×
13
D
=
16
−
52
D
=
−36
Step 2 · Find real part (p)
Use p = −b ÷ (2 × a)
The
real
part
of
both
solutions:
the
−
b
2
a
of
the
formula.
p
=
−
−4
2 × 1
p
=
4
2
p
=
2
Step 3 · Find imaginary part (q)
Use q = √(−1 × D) ÷ (2 × |a|)
√
D
=
√
−
D
×
i,
since
√
−1
=
i:
the
i
part,
taken
plus
and
minus,
so
its
size
is
positive.
q
=
√
−1 × (−36)
2 × |1|
q
=
√
36
2
q
=
6
2
x
=
4 ± 6i
2
=
2
±
3i
q
=
3
Check
Put the numbers back in. Both sides should match.
−36
=
(−4)
2
−
4
×
1
×
13
✓
2
=
−
−4
2 × 1
✓
3
=
√
−1 × (−36)
2 × |1|
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.