Let's go through these problems one by one.

Problem 1

The equation given is: $4x^2 + 2x - 6 = 0$ This is a quadratic equation of the form $ax^2 + bx + c = 0$, where $a = 4$, $b = 2$, and $c = -6$. To find the roots of this equation, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting in the values of $a$, $b$, and $c$: $x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 4 \cdot (-6)}}{2 \cdot 4}$ $x = \frac{-2 \pm \sqrt{4 + 96}}{8}$ $x = \frac{-2 \pm \sqrt{100}}{8}$ $x = \frac{-2 \pm 10}{8}$ Thus, the two solutions are:

1. $x = \frac{-2 + 10}{8} = 1$
2. $x = \frac{-2 - 10}{8} = -1.5$

So, the roots of the equation are $x = 1$ and $x = -1.5$.

Problem 2

The given equation is: $x^2 + 8x + 6y - 14 = 0$ We need to rewrite this in the standard form of a parabola equation, $(x - h)^2 = 4p(y - k)$ or $(y - k)^2 = 4p(x - h)$, and identify the vertex and the direction in which the parabola opens.

Let's isolate $y$: $x^2 + 8x - 14 = -6y$ Divide by $-6$ on both sides: $y = -\frac{1}{6}x^2 - \frac{4}{3}x + \frac{7}{3}$ This equation represents a parabola that opens downwards (since the coefficient of $x^2$ is negative).

To find the vertex, we could complete the square for the $x$-terms, but it’s sufficient here to recognize that the form of the equation is similar to $y = ax^2 + bx + c$ with $a < 0$, indicating a downward-opening parabola.

Would you like further details on any of these steps or have any additional questions?

Related Questions:

1. How do you complete the square for different types of quadratic equations?
2. Can the quadratic formula be derived from completing the square?
3. What are the different forms of the equation of a parabola, and how are they used?
4. How does the sign of the leading coefficient in a quadratic equation affect the graph's direction?
5. What are real-world applications of parabolic equations in physics and engineering?

Tip: Always check the sign of the coefficient in front of the $x^2$ term to determine the direction of a parabola's opening.