A second-degree equation that can be written in the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are real numbers and [tex]\( a \neq 0 \)[/tex], is called a quadratic equation.
Let's illustrate the solution with an example:
Consider the quadratic equation [tex]\( 2x^2 + 3x + 1 = 0 \)[/tex] where [tex]\( a = 2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = 1 \)[/tex].
To find the solutions (roots) of the quadratic equation, we can use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
First, compute the discriminant [tex]\( \Delta \)[/tex] which is given by [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[
\Delta = 3^2 - 4(2)(1) = 9 - 8 = 1
\][/tex]
The discriminant is 1. Since the discriminant is greater than 0, we have two distinct real roots. We can now find the roots.
For the first root:
[tex]\[
x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-3 + \sqrt{1}}{2(2)} = \frac{-3 + 1}{4} = \frac{-2}{4} = -0.5
\][/tex]
For the second root:
[tex]\[
x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-3 - \sqrt{1}}{2(2)} = \frac{-3 - 1}{4} = \frac{-4}{4} = -1
\][/tex]
Thus, the roots of the quadratic equation [tex]\( 2x^2 + 3x + 1 = 0 \)[/tex] are [tex]\( -0.5 \)[/tex] and [tex]\( -1 \)[/tex], and the discriminant is [tex]\( 1 \)[/tex].
