To solve the quadratic trinomial $x^2 - x - 12 = 0$, we can use the method of factoring, the quadratic formula, or by completing the square. Here, we'll start with factoring.

Step 1: Factor the Quadratic

We need to find two numbers that multiply to the constant term (-12) and add up to the coefficient of the linear term (-1).

• The constant term is $-12$.
• The coefficient of the linear term is $-1$.

We look for two numbers whose product is $-12$ and whose sum is $-1$. These numbers are $-4$ and $3$ because: $-4 \times 3 = -12 \quad \text{and} \quad -4 + 3 = -1$

Step 2: Write the Factored Form

Using the numbers $-4$ and $3$, we can write the quadratic as: $x^2 - x - 12 = (x - 4)(x + 3) = 0$

Step 3: Solve for $x$

Now, we set each factor equal to zero and solve for $x$: $x - 4 = 0 \quad \text{or} \quad x + 3 = 0$ This gives us: $x = 4 \quad \text{or} \quad x = -3$

Conclusion

The solutions to the quadratic equation $x^2 - x - 12 = 0$ are $x = 4$ and $x = -3$.

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Would you like more details on this solution or further explanation?

Here are 8 follow-up questions related to quadratic trinomials:

1. How do you solve quadratic equations using the quadratic formula?
2. What is the significance of the discriminant in quadratic equations?
3. How can completing the square be used to solve $x^2 - x - 12 = 0$?
4. What are the graphical interpretations of the roots of a quadratic equation?
5. Can the method of factoring always be used to solve quadratic equations?
6. What is the vertex form of a quadratic equation and how is it derived?
7. How do the roots of the equation $x^2 - x - 12 = 0$ relate to the graph of $y = x^2 - x - 12$?
8. How do you solve quadratic equations that do not have real roots?

Tip: When factoring quadratic trinomials, always check the product and sum of the numbers carefully to ensure the correct factors are used.