Quadratic Equations with Complex Solutions
Quadratic Equations with Complex Solutions
When solving quadratic equations, you might encounter situations where the quadratic formula requires taking the square root of a negative number. This is where complex numbers come into play.
The Discriminant and Imaginary Numbers
For any quadratic equation in the form ax2+bx+c=0, the roots are found using the quadratic formula:
x=2aâb±b2â4acââ
The expression inside the square root, b2â4ac, is called the discriminant.
- If b2â4ac>0, there are two real roots.
- If b2â4ac=0, there is one real root.
- If b2â4ac<0, there are two complex roots.
To evaluate the square root of a negative number, we use the imaginary unit i, where i2=â1 (meaning i=â1â).
Complex Conjugate Pairs
An important rule in algebra is that if a polynomial has entirely real coefficients, its complex roots will always come in conjugate pairs. This means if a+bi is a root, then aâbi must also be a root.
Example 1: Solving a Quadratic Equation
Problem: Solve x2+4x+13=0.
Solution: Here, a=1, b=4, and c=13.
First, find the discriminant: b2â4ac=42â4(1)(13)=16â52=â36
Now, plug this into the quadratic formula: x=2â4±â36ââ
Since â36â=36ââ â1â=6i, we have: x=2â4±6iâ
Divide both terms in the numerator by 2: x=â2±3i
The solutions are the complex conjugate pair â2+3i and â2â3i.
Example 2: Finding an Equation from Complex Roots
Problem: Find a quadratic equation with roots 2+3i and 2â3i.
Solution: If r1â and r2â are roots of a quadratic equation, the equation can be written as: x2â(r1â+r2â)x+(r1ââ r2â)=0
-
Find the sum of the roots: (2+3i)+(2â3i)=4
-
Find the product of the roots: (2+3i)(2â3i)=22â(3i)2=4â9i2 Since i2=â1, the product is: 4â9(â1)=4+9=13
-
Write the equation: Substitute the sum and product back into the formula: x2â4x+13=0