Sure, let's solve the quadratic equation [tex]\(6x^2 + x - 1 = 0\)[/tex] step-by-step.
### Step 1: Identify coefficients
The equation is of the form [tex]\(ax^2 + bx + c = 0\)[/tex], where:
- [tex]\(a = 6\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = -1\)[/tex]
### Step 2: Calculate the discriminant
The discriminant of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = 1^2 - 4 \cdot 6 \cdot (-1) = 1 + 24 = 25
\][/tex]
The discriminant is [tex]\(25\)[/tex].
### Step 3: Calculate the roots using the quadratic formula
The quadratic formula to find the roots of the equation [tex]\(ax^2 + bx + c = 0\)[/tex] is:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Substituting the values of [tex]\(b\)[/tex], [tex]\(\Delta\)[/tex], and [tex]\(a\)[/tex]:
[tex]\[
x = \frac{-1 \pm \sqrt{25}}{2 \cdot 6} = \frac{-1 \pm 5}{12}
\][/tex]
### Step 4: Find the individual solutions
Now we calculate the two possible values of [tex]\(x\)[/tex]:
1. For the positive root:
[tex]\[
x = \frac{-1 + 5}{12} = \frac{4}{12} = \frac{1}{3}
\][/tex]
2. For the negative root:
[tex]\[
x = \frac{-1 - 5}{12} = \frac{-6}{12} = -\frac{1}{2}
\][/tex]
### Step 5: Summarize the solutions
The solutions to the quadratic equation [tex]\(6x^2 + x - 1 = 0\)[/tex] are:
[tex]\[
x = \frac{1}{3} \quad \text{and} \quad x = -\frac{1}{2}
\][/tex]
Looking at the answer choices:
A. [tex]\(-\frac{1}{2}\)[/tex] (correct)
B. [tex]\(\frac{1}{3}\)[/tex] (correct)
C. [tex]\(\frac{1}{2}\)[/tex] (incorrect)
D. [tex]\(-\frac{1}{3}\)[/tex] (incorrect)
So, the correct solutions to [tex]\(6x^2 + x - 1 = 0\)[/tex] are [tex]\(\frac{1}{3}\)[/tex] and [tex]\(-\frac{1}{2}\)[/tex].
