Math Snap

STEP 1

1. We are given the equation $6x^4 - x^2 - 1 = 0$.
2. We need to find the real solutions of this equation.

STEP 2

1. Perform a substitution to simplify the equation.
2. Solve the resulting quadratic equation.
3. Back-substitute to find the values of $x$.
4. Determine the real solutions.

STEP 3

Perform a substitution to simplify the equation. Let $y = x^2$. Then the equation becomes:
$$6y^2 - y - 1 = 0$$

STEP 4

Solve the quadratic equation $6y^2 - y - 1 = 0$ using the quadratic formula:
The quadratic formula is given by:
$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = 6$, $b = -1$, and $c = -1$.

STEP 5

Calculate the discriminant $b^2 - 4ac$:
$$(-1)^2 - 4(6)(-1) = 1 + 24 = 25$$

STEP 6

Substitute the values into the quadratic formula:
$$y = \frac{-(-1) \pm \sqrt{25}}{2 \times 6}$$ $$y = \frac{1 \pm 5}{12}$$

STEP 7

Calculate the two possible values for $y$:
$$y = \frac{1 + 5}{12} = \frac{6}{12} = \frac{1}{2}$$ $$y = \frac{1 - 5}{12} = \frac{-4}{12} = -\frac{1}{3}$$

STEP 8

Back-substitute to find the values of $x$.
For $y = \frac{1}{2}$:
$$x^2 = \frac{1}{2}$$ $$x = \pm \sqrt{\frac{1}{2}}$$ $$x = \pm \frac{\sqrt{2}}{2}$$ For $y = -\frac{1}{3}$:
Since $x^2 = -\frac{1}{3}$ has no real solutions (as a square cannot be negative), we discard this solution.

Determine the real solutions. The real solutions are:
$$x = \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$$ The solution set is:
A. $\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$