Find the roots of the equation below.
[tex]\[ x^2 - 6x + 12 = 0 \][/tex]

A. [tex]\(-3 \pm \sqrt{3}\)[/tex]

B. [tex]\(3 \pm \sqrt{3}\)[/tex]

C. [tex]\(3 \pm N 3\)[/tex]

D. [tex]\(-3 \pm N 3\)[/tex]



Answer :

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]