A quadratic equation can either have two real roots (zeros), a repeated real root or no real roots
This depends on the location of the graph of the quadratic with respect to the x-axis
If a quadratic equation has no real roots we would previously have stated that it has no real solutions
The quadratic equation will have a negative discriminant
This means taking the square root of a negative number
Complex numbers provide solutions for quadratic equations that have no real roots
How do we solve a quadratic equation when it has complex roots?
If a quadratic equation takes the form ax2+ bx + c = 0 it can be solved by either using the quadratic formula or completing the square
If a quadratic equation takes the form ax2+ b = 0 it can be solved by rearranging
The property i = √-1 is used
If the coefficients of the quadratic are real then the complex roots will occur in complex conjugate pairs
If z = m + ni(n ≠ 0) is a root of a quadratic with real coefficients then z* = m - ni is also a root
The real part of the solutions will have the same value as the x coordinate of the turning point on the graph of the quadratic
When the coefficients of the quadratic equation are non-real, the solutions will not be complex conjugates
To solve these you can use the quadratic formula
How do we factorise a quadratic equation if it has complex roots?
If we are given a quadratic equation in the form az2+ bz + c = 0, where a, b, and c ∈ ℝ, a ≠ 0 we can use its complex roots to write it in factorised form
Use the quadratic formula to find the two roots, z = p + qi and z*= p - qi
This means that z – (p + qi) and z – (p – qi) must both be factors of the quadratic equation
Therefore we can write az2+ bz + c = (z – (p + qi))( z – (p - qi))
This can be rearranged into the form (z – p – qi)(z – p + qi)
How do we find a quadratic equation when given a complex root?
If we are given a complex root in the form z= p + qi we can find the quadratic equation in the form az2+ bz + c = 0, where a, b, and c ∈ ℝ, a ≠ 0
We know that the second root must be z*= p - qi
This means that z – (p + qi) and z – (p – qi) must both be factors of the quadratic equation
Therefore we can write az2+ bz + c = (z – (p + qi))( z – (p - qi))
Rewriting this as ((z – p)– qi))((z – p) + qi)) makes expanding easier
Expanding this gives the quadratic equation z2 – 2pz +(p2+ q2)
a = 1
b = -2p
c = p2+ q2
This demonstrates the important property (x – z)(x – z*) = x2- 2Re(z)x + |z|2
Exam Tip
Worked Example
a) Solve the quadratic equation z2 - 2z + 5 = 0 and hence, factorise z2 - 2z + 5.
b) Given that one root of a quadratic equation is z = 2 – 3i, find the quadratic equation in the form az2+ bz + c = 0, where a, b, and c ∈ ℝ, a ≠ 0.
Complex Roots of Polynomials
How many roots should a polynomial have?
We know that every quadratic equation has two roots (not necessarily distinct or real)
This is a particular case of a more general rule:
Every polynomial equation, with real coefficients, of degree n has n roots
The n roots are not necessarily all distinct and therefore we need to count any repeated roots that may occur individually
From the above rule we can state the following:
3 real roots
Or 1 real root and a complex conjugate pair
4 real roots
2 real and 2 nonreal (a complex conjugate pair)
4 nonreal (two complex conjugate pairs)
When a polynomial of any degree has one complex root it will always also have the complex conjugate as a root
How do we solve a cubic equation with complex roots?
Steps to solve a cubic equation with complex roots
If we are told thatp + qi is a root, then we know p - qi is also a root
This means that z – (p + qi) and z – (p – qi) must both be factors of the cubic equation
Multiplying the above factors together gives us a quadratic factor of the form (Az2 + Bz + C)
We need to find the third factor (z - α)
From there either
Expand and compare coefficients to find
Or use polynomial division to find the factor (z - α)
Finally, write your three roots clearly
How do we solve a polynomial of any degree with complex roots?
When asked to find the roots of any polynomial when we are given one, we use almost the same method as for a cubic equation
State the initial root and its conjugate and write their factors as a quadratic factor (as above) we will have two unknown roots to find, write these as factors (z - α) and (z - β)
The unknown factors also form a quadratic factor (z - α)(z - β)
Then continue with the steps from above, either comparing coefficients or using polynomial division
If using polynomial division, then solve the quadratic factor you get to find the roots α and β
How do we solve polynomial equations with unknown coefficients?
Steps to find unknown variables in a given equation when given a root:
Substitute the given root p + qi into the equation f(z) = 0
Expand and group together the real and imaginary parts (these expressions will contain our unknown values)
Solve as simultaneous equations to find the unknowns
Substitute the values into the original equation
From here continue using the previously described methods for finding other roots for the polynomial
How do we factorise a polynomial when given a complex root?
If we are given a root of a polynomial of any degree in the form z = p + qi
We know that the complex conjugate, z*= p – qi is another root
We can write (z – (p + qi)) and ( z – (p - qi)) as two linear factors
Or rearrange into one quadratic factor
This can be multiplied out with another factor to find further factors of the polynomial
For higher order polynomials more than one root may be given
If the further given root is complex then its complex conjugate will also be a root
This will allow you to find further factors
Exam Tip
You can speed up multiplying two complex conjugate factors together by
rewrite (z – (p + qi))(z – (p - qi)) as ((z – p) -qi))( (z – p) +qi))
Then ((z – p) -qi))( (z – p) +qi)) = (z - p)2 - (qi)2 = (z - p)2 + q2
If you are working on a calculator paper read the question carefully to see how much of the working needs to be shown but always remember to use your GDC to check your working where you can
Worked Example
Given that one root of a polynomial p(x) = z3 + z2 –7z +65is 2 – 3i, find the other roots.