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=2a−b±b2−4ac
The expression inside the square root, b2−4ac, is called the discriminant.
- If b2−4ac>0, there are two real roots.
- If b2−4ac=0, there is one real root.
- If b2−4ac<0, there are two complex roots.
To evaluate the square root of a negative number, we use the imaginary unit i, where i2=−1 (meaning i=−1).
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 a−bi 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: b2−4ac=42−4(1)(13)=16−52=−36
Now, plug this into the quadratic formula: x=2−4±−36
Since −36=36⋅−1=6i, we have: x=2−4±6i
Divide both terms in the numerator by 2: x=−2±3i
The solutions are the complex conjugate pair −2+3i and −2−3i.
Example 2: Finding an Equation from Complex Roots
Problem: Find a quadratic equation with roots 2+3i and 2−3i.
Solution: If r1 and r2 are roots of a quadratic equation, the equation can be written as: x2−(r1+r2)x+(r1⋅r2)=0
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Find the sum of the roots: (2+3i)+(2−3i)=4
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Find the product of the roots: (2+3i)(2−3i)=22−(3i)2=4−9i2 Since i2=−1, the product is: 4−9(−1)=4+9=13
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Write the equation: Substitute the sum and product back into the formula: x2−4x+13=0