Certainly! To solve the quadratic equation [tex]\(3x^2 - 4\sqrt{3}x + 4 = 0\)[/tex], let's follow the quadratic formula step by step.
The quadratic formula to find the roots of an equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are:
[tex]\[ a = 3, \quad b = -4\sqrt{3}, \quad c = 4 \][/tex]
Let's solve for the discriminant first:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ b^2 = (-4\sqrt{3})^2 = 16 \cdot 3 = 48 \][/tex]
[tex]\[ 4ac = 4 \cdot 3 \cdot 4 = 48 \][/tex]
[tex]\[ \text{Discriminant} = 48 - 48 = 0 \][/tex]
Since the discriminant is zero, we know that the quadratic equation has a single (repeated) real root, meaning both roots are the same.
Next, we find the root using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{0}}{2a} = \frac{-b}{2a} \][/tex]
Substituting the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ x = \frac{-(-4\sqrt{3})}{2 \cdot 3} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} \][/tex]
Therefore, the single repeated solution to the quadratic equation [tex]\(3x^2 - 4\sqrt{3}x + 4 = 0\)[/tex] is:
[tex]\[ x = \frac{2\sqrt{3}}{3} \][/tex]
