Quadratic Equations With Complex Roots Revision Notes for Leaving Cert Mathematics

Quadratic Equations With Complex Roots

When do quadratic equations have complex roots?

Complex roots occur when a quadratic equation cannot be factorised using real numbers. This happens when we need to use the quadratic formula to solve the equation.

For any quadratic equation in the form $ax^2 + bx + c = 0$ , we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Key rule: If $b^2 - 4ac$ is negative (less than zero), then the roots will be complex numbers.

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The expression $b^2 - 4ac$ is called the discriminant. When the discriminant is negative, we get complex roots because we need to find the square root of a negative number.

Understanding complex roots

When a quadratic equation has complex roots, we typically use $z$ as the variable instead of $x$ . The roots are always expressed in the standard form $a + bi$ , where:

$a$ is the real part
$b$ is the coefficient of the imaginary part
$i$ represents $\sqrt{-1}$

Important property: conjugate pairs

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Complex roots always come in conjugate pairs. This means if one root is $a + bi$ , then the other root must be $a - bi$ .

For example:

If $2 + 3i$ is a root, then $2 - 3i$ is also a root
If $-3 + i$ is a root, then $-3 - i$ is also a root

Worked example 1: solving a quadratic with complex roots

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Worked Example: Solving for Complex Roots

Solve: $z^2 + 6z + 10 = 0$

Step 1: Identify the coefficients

$a = 1$
$b = 6$
$c = 10$

Step 2: Check the discriminant $b^2 - 4ac = 6^2 - 4(1)(10) = 36 - 40 = -4$

Since the discriminant is negative, we will get complex roots.

Step 3: Apply the quadratic formula $z = \frac{-6 \pm \sqrt{-4}}{2(1)} = \frac{-6 \pm \sqrt{-4}}{2}$

Step 4: Simplify the square root of the negative number Since $\sqrt{-4} = \sqrt{4} \times \sqrt{-1} = 2i$

$z = \frac{-6 \pm 2i}{2} = \frac{-6}{2} \pm \frac{2i}{2} = -3 \pm i$

Therefore: $z = -3 + i$ or $z = -3 - i$

Worked example 2: finding unknown constants

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Worked Example: Finding Unknown Constants

Problem: If $z = 3 + i$ is a root of the equation $z^2 - 6z + k = 0$ , find the value of $k$ and write down the other root.

Step 1: Since $z = 3 + i$ is a root, it satisfies the equation

Substitute $(3 + i)$ for $z$ in $z^2 - 6z + k = 0$ :

$(3 + i)^2 - 6(3 + i) + k = 0$

Step 2: Expand $(3 + i)^2$ $(3 + i)^2 = 9 + 6i + i^2 = 9 + 6i - 1 = 8 + 6i$

Step 3: Substitute and simplify $(8 + 6i) - 6(3 + i) + k = 0$ $8 + 6i - 18 - 6i + k = 0$ $-10 + k = 0$ $k = 10$

Step 4: Find the other root Since complex roots come in conjugate pairs, the other root is $3 - i$ .

Worked example 3: standard quadratic with complex roots

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Worked Example: Standard Form Solution

Solve: $z^2 + 4z + 5 = 0$

Step 1: Identify coefficients: $a = 1$ , $b = 4$ , $c = 5$

Step 2: Apply the quadratic formula $z = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}$

Step 3: Simplify $z = \frac{-4 \pm 2i}{2} = -2 \pm i$

Therefore: $z = -2 + i$ or $z = -2 - i$

Key exam tips

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Essential Exam Strategies:

Always check if the discriminant $b^2 - 4ac$ is negative before expecting complex roots
Remember that $\sqrt{-1} = i$ and $i^2 = -1$
Complex roots always come in conjugate pairs - if you find one, you automatically know the other
When substituting a complex number into an equation, be careful with $(a + bi)^2$ expansion
Express your final answers in the form $a + bi$ or $a - bi$

Remember!

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Key Points to Remember:

Complex roots occur when the discriminant $b^2 - 4ac < 0$
Always use the quadratic formula when you cannot factorise and suspect complex roots
Complex roots come in conjugate pairs - change the sign of the imaginary part to find the partner
Standard form is $a + bi$ where $a$ is real and $bi$ is imaginary
Check your work by substituting back into the original equation

Quadratic Equations With Complex Roots (Leaving Cert Mathematics): Revision Notes