Let's solve the equation step by step:
Given equation:
[ 3x^2 - x = 1 - 3x^2 + x ]
First, let's move all terms to one side of the equation:
[
3x^2 - x + 3x^2 - x - 1 = 0
]
Combining like terms, we get:
[
6x^2 - 2x - 1 = 0
]
Now, we can apply the quadratic formula, which is:
[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
]
In our equation ( a = 6 ), ( b = -2 ), and ( c = -1 ).
Now plug in the values:
- Calculate the discriminant ( b^2 - 4ac ):
[
(-2)^2 - 4 \cdot 6 \cdot (-1) = 4 + 24 = 28
]
- Now substitute into the quadratic formula:
[
x = \frac{-(-2) \pm \sqrt{28}}{2 \cdot 6}
]
[
x = \frac{2 \pm \sqrt{28}}{12}
]
- Simplify ( \sqrt{28} = 2\sqrt{7} ):
[
x = \frac{2 \pm 2\sqrt{7}}{12}
]
[
x = \frac{1 \pm \sqrt{7}}{6}
]
Thus, the solutions to the equation are:
[
x = \frac{1 + \sqrt{7}}{6} \quad \text{and} \quad x = \frac{1 - \sqrt{7}}{6}
]
