Find the solutions to the equation below. Check all that apply.

[tex]6x^2 + 5x - 4 = 0[/tex]

A. [tex]x = 4[/tex]
B. [tex]x = -\frac{4}{3}[/tex]
C. [tex]x = \frac{1}{2}[/tex]
D. [tex]x = -2[/tex]
E. [tex]x = -\frac{1}{3}[/tex]
F. [tex]x = \frac{1}{3}[/tex]



Answer :

To find the solutions to the quadratic equation [tex]\(6x^2 + 5x - 4 = 0\)[/tex], we can use the quadratic formula, which states that for any quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], the solutions for [tex]\(x\)[/tex] can be found using:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Here, our equation is [tex]\(6x^2 + 5x - 4 = 0\)[/tex]. We can identify the coefficients as follows:
- [tex]\(a = 6\)[/tex]
- [tex]\(b = 5\)[/tex]
- [tex]\(c = -4\)[/tex]

Plugging these values into the quadratic formula, we get:

[tex]\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 6 \cdot (-4)}}{2 \cdot 6} \][/tex]

Let's calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]) first:

[tex]\[ b^2 - 4ac = 5^2 - 4 \cdot 6 \cdot (-4) = 25 + 96 = 121 \][/tex]

So, we have:

[tex]\[ x = \frac{-5 \pm \sqrt{121}}{12} = \frac{-5 \pm 11}{12} \][/tex]

This results in two solutions:

1. For the positive root:

[tex]\[ x = \frac{-5 + 11}{12} = \frac{6}{12} = \frac{1}{2} \][/tex]

2. For the negative root:

[tex]\[ x = \frac{-5 - 11}{12} = \frac{-16}{12} = -\frac{4}{3} \][/tex]

Therefore, the solutions to the equation [tex]\(6x^2 + 5x - 4 = 0\)[/tex] are [tex]\(x = \frac{1}{2}\)[/tex] and [tex]\(x = -\frac{4}{3}\)[/tex].

From the given options:
- [tex]\(x = 4\)[/tex]
- [tex]\(x = -\frac{4}{3}\)[/tex]
- [tex]\(x = \frac{1}{2}\)[/tex]
- [tex]\(x = -2\)[/tex]
- [tex]\(x = -\frac{1}{3}\)[/tex]
- [tex]\(x = \frac{1}{3}\)[/tex]

We see that the correct solutions that match the options are:
- [tex]\(x = \frac{1}{2}\)[/tex]
- [tex]\(x = -\frac{4}{3}\)[/tex]

Thus, the correct answers are:
B. [tex]\(x = -\frac{4}{3}\)[/tex]
C. [tex]\(x = \frac{1}{2}\)[/tex]