Edexcel IAL - Further Pure Mathematics 1- 1.4 Solving Quadratic Equations with Complex Roots- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 1- 1.4 Solving Quadratic Equations with Complex Roots -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 1- 1.4 Solving Quadratic Equations with Complex Roots -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 1.4 Solving Quadratic Equations with Complex Roots
Complex Solutions of Quadratic Equations with Real Coefficients
A quadratic equation with real coefficients has the form:
\( ax^2 + bx + c = 0 \), where \( a,b,c \) are real and \( a \neq 0 \).
If the discriminant \( b^2 – 4ac \) is negative, the equation has no real solutions. Instead, it has two complex solutions.
Quadratic Formula
The solutions of any quadratic equation are given by:
\( x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
If \( b^2 – 4ac < 0 \), write:
\( \sqrt{b^2 – 4ac} = \sqrt{-1}\sqrt{4ac – b^2} = i\sqrt{4ac – b^2} \)
So the roots are complex numbers.
Conjugate Pair Property
If the coefficients \( a,b,c \) are real and one root is \( p + iq \), then the other root is \( p – iq \).
This means complex roots always occur in conjugate pairs.
Example
Solve \( x^2 + 4x + 13 = 0 \).
▶️ Answer / Explanation
Discriminant:
\( b^2 – 4ac = 16 – 52 = -36 \)
\( \sqrt{-36} = 6i \)
\( x = \dfrac{-4 \pm 6i}{2} = -2 \pm 3i \)
Example
Solve \( 2x^2 – 4x + 5 = 0 \).
▶️ Answer / Explanation
\( b^2 – 4ac = 16 – 40 = -24 \)
\( \sqrt{-24} = \sqrt{24}i = 2\sqrt{6}i \)
\( x = \dfrac{4 \pm 2\sqrt{6}i}{4} = 1 \pm \dfrac{\sqrt{6}}{2}i \)
Example
Find the quadratic equation with real coefficients whose roots are \( 3 + 2i \) and \( 3 – 2i \).
▶️ Answer / Explanation
Let the roots be \( \alpha = 3 + 2i \), \( \beta = 3 – 2i \).
Sum of roots:
\( \alpha + \beta = 6 \)
Product of roots:
\( \alpha\beta = (3 + 2i)(3 – 2i) = 9 + 4 = 13 \)
The quadratic equation is:
\( x^2 – 6x + 13 = 0 \)