A second-degree probable equation typically refers to a quadratic equation, which is of the form:

$ax^2 + bx + c = 0$

Where:

• $a$, $b$, and $c$ are constants (real numbers),
• $x$ represents the unknown variable,
• $a \neq 0$ (since if $a = 0$, it would become a linear equation, not quadratic).

Solving the Quadratic Equation

To find the solutions (roots) of the equation, we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where:

• $b^2 - 4ac$ is called the discriminant.
• If $b^2 - 4ac > 0$, there are two distinct real solutions.
• If $b^2 - 4ac = 0$, there is one real solution (a repeated root).
• If $b^2 - 4ac < 0$, there are no real solutions (but two complex solutions).

Example:

If we have the equation: $2x^2 + 3x - 5 = 0$

Here, $a = 2$, $b = 3$, and $c = -5$.

Using the quadratic formula: $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)}$ $x = \frac{-3 \pm \sqrt{9 + 40}}{4}$ $x = \frac{-3 \pm \sqrt{49}}{4}$ $x = \frac{-3 \pm 7}{4}$

So, the two solutions are: $x = \frac{-3 + 7}{4} = \frac{4}{4} = 1$ and $x = \frac{-3 - 7}{4} = \frac{-10}{4} = -2.5$

Thus, the solutions are $x = 1$ and $x = -2.5$.

Would you like more detailed steps or another example?

Follow-up Questions:

1. What is the discriminant, and how does it affect the number of solutions?
2. How do you solve quadratic equations by factoring?
3. How can you solve a quadratic equation using completing the square?
4. What are complex roots in quadratic equations, and when do they occur?
5. Can a quadratic equation have irrational solutions? If so, how?

Tip: To quickly check the nature of the roots (real or complex), calculate the discriminant $b^2 - 4ac$ first!