Certainly! To find the roots of the quadratic equation [tex]\( x^2 - 6x + 12 = 0 \)[/tex], we will use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the coefficients of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
From the given equation [tex]\( x^2 - 6x + 12 = 0 \)[/tex], we can identify that:
[tex]\[
a = 1, \quad b = -6, \quad c = 12
\][/tex]
Step-by-step, we calculate as follows:
1. Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
\Delta = (-6)^2 - 4(1)(12)
\][/tex]
[tex]\[
\Delta = 36 - 48
\][/tex]
[tex]\[
\Delta = -12
\][/tex]
2. Calculate the roots:
Since the discriminant [tex]\( \Delta = -12 \)[/tex] is negative, the roots will be complex numbers. We use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Substitute [tex]\( \Delta = -12 \)[/tex], [tex]\( a = 1 \)[/tex], and [tex]\( b = -6 \)[/tex]:
[tex]\[
x = \frac{-(-6) \pm \sqrt{-12}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{6 \pm \sqrt{-12}}{2}
\][/tex]
We know that [tex]\( \sqrt{-12} = \sqrt{12} \cdot i \)[/tex], where [tex]\( i \)[/tex] is the imaginary unit ([tex]\( i^2 = -1 \)[/tex]). Further, [tex]\( \sqrt{12} \)[/tex] can be simplified to [tex]\( 2\sqrt{3} \)[/tex]:
[tex]\[
\sqrt{-12} = 2\sqrt{3} \cdot i
\][/tex]
Substituting this into the formula gives:
[tex]\[
x = \frac{6 \pm 2\sqrt{3} \cdot i}{2}
\][/tex]
[tex]\[
x = 3 \pm \sqrt{3} \cdot i
\][/tex]
Hence, the roots of the equation [tex]\( x^2 - 6x + 12 = 0 \)[/tex] are:
[tex]\[
x = 3 \pm \sqrt{3}i
\][/tex]
