To find the roots of the quadratic equation [tex]\(6x^2 + 5x - 4 = 0\)[/tex], we can use the quadratic formula, which is given by:
[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].
For the given quadratic equation:
[tex]\[ a = 6 \][/tex]
[tex]\[ b = 5 \][/tex]
[tex]\[ c = -4 \][/tex]
First, we need to calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 5^2 - 4 \cdot 6 \cdot (-4) \][/tex]
[tex]\[ \Delta = 25 + 96 \][/tex]
[tex]\[ \Delta = 121 \][/tex]
The discriminant is 121. Now, we substitute [tex]\(\Delta\)[/tex], [tex]\(a\)[/tex], and [tex]\(b\)[/tex] into the quadratic formula to find the roots:
[tex]\[ x = \frac{-5 \pm \sqrt{121}}{2 \cdot 6} \][/tex]
[tex]\[ x = \frac{-5 \pm 11}{12} \][/tex]
This gives us two solutions:
1. The first root:
[tex]\[ x_1 = \frac{-5 + 11}{12} \][/tex]
[tex]\[ x_1 = \frac{6}{12} \][/tex]
[tex]\[ x_1 = \frac{1}{2} \][/tex]
2. The second root:
[tex]\[ x_2 = \frac{-5 - 11}{12} \][/tex]
[tex]\[ x_2 = \frac{-16}{12} \][/tex]
[tex]\[ x_2 = -\frac{4}{3} \][/tex]
Therefore, the roots of the quadratic equation [tex]\(6x^2 + 5x - 4 = 0\)[/tex] are:
[tex]\[ x = \frac{1}{2} \quad \text{and} \quad x = -\frac{4}{3} \][/tex]
To match the given answer options, we can express [tex]\(-\frac{4}{3}\)[/tex] as a mixed number:
[tex]\[ -\frac{4}{3} = -1 \frac{1}{3} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{\frac{1}{2} \text{ and } -1 \frac{1}{3}} \][/tex]
