Question
why are the solutions of a quadratic equation with a negative discriminant complex solutions?
Answers
Answered by
GPT-4o mini
A quadratic equation can be expressed in the standard form:
\[
ax^2 + bx + c = 0
\]
where \( a \), \( b \), and \( c \) are constants, and \( a \neq 0 \). To find the solutions (or roots) of this equation, we use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{D}}{2a}
\]
where \( D \) is the discriminant, given by:
\[
D = b^2 - 4ac
\]
The value of the discriminant \( D \) determines the nature of the roots of the quadratic equation:
1. If \( D > 0 \), there are two distinct real solutions.
2. If \( D = 0 \), there is exactly one real solution (a repeated root).
3. If \( D < 0 \), there are no real solutions, and the solutions are complex.
When \( D < 0 \), the term under the square root in the quadratic formula, \( \sqrt{D} \), becomes the square root of a negative number. In mathematics, the square root of a negative number is not a real number, but a complex number.
We express the square root of a negative number using the imaginary unit \( i \), where:
\[
i = \sqrt{-1}
\]
Thus, if \( D < 0 \), we can express it as:
\[
D = -k \quad \text{for some } k > 0
\]
Then:
\[
\sqrt{D} = \sqrt{-k} = \sqrt{k}i
\]
Substituting this back into the quadratic formula gives:
\[
x = \frac{-b \pm \sqrt{-k}i}{2a} = \frac{-b \pm \sqrt{k}i}{2a}
\]
This results in two complex solutions, which can be expressed as:
\[
x = \frac{-b}{2a} \pm \frac{\sqrt{k}}{2a}i
\]
In summary, when the discriminant is negative, the roots become complex due to the square root of a negative number resulting in the imaginary unit \( i \), leading to solutions that are not only real numbers but rather complex numbers.
\[
ax^2 + bx + c = 0
\]
where \( a \), \( b \), and \( c \) are constants, and \( a \neq 0 \). To find the solutions (or roots) of this equation, we use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{D}}{2a}
\]
where \( D \) is the discriminant, given by:
\[
D = b^2 - 4ac
\]
The value of the discriminant \( D \) determines the nature of the roots of the quadratic equation:
1. If \( D > 0 \), there are two distinct real solutions.
2. If \( D = 0 \), there is exactly one real solution (a repeated root).
3. If \( D < 0 \), there are no real solutions, and the solutions are complex.
When \( D < 0 \), the term under the square root in the quadratic formula, \( \sqrt{D} \), becomes the square root of a negative number. In mathematics, the square root of a negative number is not a real number, but a complex number.
We express the square root of a negative number using the imaginary unit \( i \), where:
\[
i = \sqrt{-1}
\]
Thus, if \( D < 0 \), we can express it as:
\[
D = -k \quad \text{for some } k > 0
\]
Then:
\[
\sqrt{D} = \sqrt{-k} = \sqrt{k}i
\]
Substituting this back into the quadratic formula gives:
\[
x = \frac{-b \pm \sqrt{-k}i}{2a} = \frac{-b \pm \sqrt{k}i}{2a}
\]
This results in two complex solutions, which can be expressed as:
\[
x = \frac{-b}{2a} \pm \frac{\sqrt{k}}{2a}i
\]
In summary, when the discriminant is negative, the roots become complex due to the square root of a negative number resulting in the imaginary unit \( i \), leading to solutions that are not only real numbers but rather complex numbers.