The equation (4x^2 – 5x – 12 = 0) represents a quadratic function that can often stump those encountering it for the first time. **By applying the quadratic formula, anyone can find the roots of this equation and understand its behavior on a graph.** This fundamental skill is essential for students and professionals alike, as it opens doors to advanced topics in math and science.

Quadratics are more than just numbers; they describe various real-world scenarios, from projectile motion to business optimizations. Solving (4x^2 – 5x – 12 = 0) will not only provide specific solutions but also demonstrate the workings of quadratic functions in practice.

Ultimately, mastering the techniques to solve such equations can greatly enhance mathematical confidence and competence. Readers are invited to explore the solution process and discover how easily they can tackle similar challenges.

## Quadratic Equations Overview

Quadratic equations are polynomial equations of the form (ax^2 + bx + c = 0), where (a), (b), and (c) are constants. The variable (x) represents an unknown value. The leading coefficient (a) must be non-zero for it to be classified as quadratic.

These equations can have zero, one, or two real solutions. The nature of the solutions is determined by the discriminant, represented as (D = b^2 – 4ac).

### Discriminant Cases

**Positive Discriminant ((D > 0))**: Two distinct real solutions.**Zero Discriminant ((D = 0))**: Exactly one real solution (a repeated root).**Negative Discriminant ((D < 0))**: No real solutions (two complex solutions).

Quadratic equations can be solved through various methods including:

**Factoring**: Expressing the quadratic as a product of two binomials.**Completing the Square**: Rearranging the equation into a perfect square format.**Quadratic Formula**: Using the formula (x = \frac{-b \pm \sqrt{D}}{2a}) to find solutions.

Quadratics are widely used in various fields such as physics, engineering, and finance, making them an essential part of mathematical studies. They also graph as parabolas, which exhibit symmetry about a vertical line known as the axis of symmetry.

## Solving the Quadratic Equation 4x^2 – 5x – 12 = 0

This section addresses three methods to solve the quadratic equation (4x^2 – 5x – 12 = 0). Readers will learn how to use the factoring method, the quadratic formula, and the technique of completing the square.

### Factoring Method

To solve (4x^2 – 5x – 12 = 0) using factoring, the first step is to rewrite the expression. This equation can be factored into two binomials.

The goal is to find two numbers that multiply to (4 \times -12 = -48) and add to (-5). The numbers (-8) and (3) satisfy these conditions. The equation can be expressed as:

[ 4x^2 – 8x + 3x – 12 = 0 ]

Next, factor by grouping:

[ 4x(x – 2) + 3(x – 2) = 0 ]

The factored form is:

[ (4x + 3)(x – 2) = 0 ]

Setting each factor to zero gives (4x + 3 = 0) or (x – 2 = 0). Therefore, the solutions are:

[ x = -\frac{3}{4} \quad \text{and} \quad x = 2 ]

### Quadratic Formula Application

The quadratic formula, given by

[ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} ]

can be applied directly to find the roots. Here, (a = 4), (b = -5), and (c = -12).

First, calculate the discriminant:

[ b^2 – 4ac = (-5)^2 – 4(4)(-12) = 25 + 192 = 217 ]

Next, substitute (a), (b), and the discriminant into the formula:

[ x = \frac{5 \pm \sqrt{217}}{8} ]

This leads to two solutions:

[ x = \frac{5 + \sqrt{217}}{8} \quad \text{and} \quad x = \frac{5 – \sqrt{217}}{8} ]

These results provide the exact roots of the equation.

### Completing the Square Technique

To solve (4x^2 – 5x – 12 = 0) by completing the square, start by dividing the equation by (4):

[ x^2 – \frac{5}{4}x – 3 = 0 ]

Rearranging yields:

[ x^2 – \frac{5}{4}x = 3 ]

Next, find the term to complete the square. Take (\left(\frac{-5/4}{2}\right)^2 = \frac{25}{64}) and add it to both sides:

[ x^2 – \frac{5}{4}x + \frac{25}{64} = 3 + \frac{25}{64} ]

Convert (3) to a fraction with a denominator of (64):

[ 3 = \frac{192}{64} ]

This gives:

[ x^2 – \frac{5}{4}x + \frac{25}{64} = \frac{217}{64} ]

Now, write the left-hand side as a square:

[ \left(x – \frac{5}{8}\right)^2 = \frac{217}{64} ]

Taking the square root of both sides provides:

[ x – \frac{5}{8} = \pm \frac{\sqrt{217}}{8} ]

The final solutions are:

[ x = \frac{5 \pm \sqrt{217}}{8} ]

**How to do a quadratic equation formula?**

**The quadratic formula is a general formula used to solve any quadratic equation.**

### The Formula

The quadratic formula is:

```
x = (-b ± √(b² - 4ac)) / (2a)
```

Where:

`a`

,`b`

, and`c`

are the coefficients of the quadratic equation in the standard form`ax² + bx + c = 0`

.

### Steps to Solve

**Write the equation in standard form:**Make sure the equation is in the form`ax² + bx + c = 0`

.**Identify the values of a, b, and c:**Determine the coefficients of the quadratic equation.**Substitute the values into the formula:**Replace`a`

,`b`

, and`c`

in the formula with their corresponding values.**Simplify:**Perform the calculations within the formula, following the order of operations (PEMDAS/BODMAS).**Find the two solutions:**The ± symbol indicates that there are two possible solutions. Calculate both by using the plus and minus signs separately.

### Example

Let’s solve the quadratic equation: `2x² + 5x - 3 = 0`

**Standard form:**The equation is already in standard form.**Identify a, b, and c:**`a = 2`

`b = 5`

`c = -3`

**Substitute into the formula:**`x = (-5 ± √(5² - 4*2*(-3))) / (2*2)`

**Simplify:**`x = (-5 ± √(49)) / 4`

`x = (-5 ± 7) / 4`

**Find the solutions:**`x₁ = (-5 + 7) / 4 = 2 / 4 = 1/2`

`x₂ = (-5 - 7) / 4 = -12 / 4 = -3`

Therefore, the solutions to the equation `2x² + 5x - 3 = 0`

are `x = 1/2`

and `x = -3`

.

**How do you write the equation of a quadratic equation?**

## Writing the Equation of a Quadratic Equation

**Standard Form of a Quadratic Equation:**

A quadratic equation is typically written in the standard form:

```
ax² + bx + c = 0
```

Where:

`a`

,`b`

, and`c`

are constants (numbers)`x`

is the variable`a`

cannot be equal to 0 (if it were, the equation would be linear, not quadratic)

### Examples of Quadratic Equations:

`2x² + 5x - 3 = 0`

`-x² + 4 = 0`

`x² - 9 = 0`

### Creating Your Own Quadratic Equation

To create a quadratic equation, simply assign values to `a`

, `b`

, and `c`

. For example:

- Let
`a = 3`

,`b = -2`

, and`c = 1`

. The equation would be:`3x² - 2x + 1 = 0`

### Other Forms of Quadratic Equations

While the standard form is common, there are other forms:

**Vertex Form:**`y = a(x - h)² + k`

- Used when you know the vertex (h, k) of the parabola

**Factored Form:**`y = a(x - r₁)(x - r₂)`

- Used when you know the x-intercepts (r₁ and r₂)

**How to find the equation of a quadratic function?**

## Finding the Equation of a Quadratic Function

**Understanding the Basics**

A quadratic function is generally expressed as:

```
y = ax² + bx + c
```

Where:

`a`

,`b`

, and`c`

are constants`x`

is the independent variable`y`

is the dependent variable

To find the equation, you typically need specific information about the function. Here are common scenarios:

### 1. Given Three Points

If you have three points (x₁, y₁), (x₂, y₂), and (x₃, y₃) that lie on the parabola, you can set up a system of three equations:

```
y₁ = ax₁² + bx₁ + c
y₂ = ax₂² + bx₂ + c
y₃ = ax₃² + bx₃ + c
```

Solving this system of equations will give you the values of `a`

, `b`

, and `c`

, allowing you to write the equation.

### 2. Given the Vertex and Another Point

If you know the vertex (h, k) and another point (x, y) on the parabola, you can use the vertex form of the quadratic equation:

```
y = a(x - h)² + k
```

Substitute the values of `h`

, `k`

, `x`

, and `y`

into the equation and solve for `a`

.

### 3. Given the x-intercepts and Another Point

If you know the x-intercepts (p, 0) and (q, 0) and another point (x, y) on the parabola, you can use the factored form of the quadratic equation:

```
y = a(x - p)(x - q)
```

Substitute the values of `p`

, `q`

, `x`

, and `y`

into the equation and solve for `a`

.

### Additional Considerations

**Graphing the function**can provide visual insights into its shape and key points.**Using technology**(like graphing calculators or software) can simplify calculations and provide accurate results.

**Example:** Find the equation of the quadratic function passing through the points (1, 2), (2, 3), and (-1, 6).

Using the system of equations method, we get:

```
2 = a + b + c
3 = 4a + 2b + c
6 = a - b + c
```

Solving this system, we find a = 1, b = -1, and c = 2.

Therefore, the equation is:

```
y = x² - x + 2
```

**Would you like to try an example?** Please provide the given information (points, vertex, x-intercepts, etc.), and I can guide you through the process.

**Conclusion**

Sure, I’ve been following along. Let’s conclude the process of solving the quadratic equation:$4x_{2}−5x−12=0$

We used the quadratic formula to solve for the roots, which are the values of x that make the expression equal to zero. The quadratic formula is a general formula that works for any quadratic equation.

Steps to solve:

- Identify the coefficients a, b, and c of the quadratic equation. In this case, a = 4, b = -5, and c = -12.
- Substitute these coefficients into the quadratic formula:$x=ab±b−ac $
- Simplify the expression under the radical (the part inside the square root). This is called the discriminant. In this case, the discriminant is (-5)^2 – 4 * 4 * (-12) = 101.
- Because the discriminant is not a perfect square, the quadratic formula reduces to $\frac{-b \pm \sqrt{101}}{2a} $, which means there are two irrational roots.
- Calculate the two roots by substituting the values of a, b, and the discriminant into the formula.

The solutions to the equation are:$x1=2.4663649828320295$$x2=−1.2163649828320293$

These solutions were found using the quadratic formula, which is a general purpose method for solving quadratic equations.