Solving Quadratic Equations: A Step-by-Step Guide to 4x^2 – 5x – 12 = 0

Welcome to our comprehensive guide on solving quadratic equations! If you’re familiar with the equation 4x^2 – 5x – 12 = 0, but find yourself scratching your head trying to figure out how to solve it, you’ve come to the right place. So grab a pen and paper (or fire up that calculator) as we dive into the exciting world of quadratic equations and unravel the mystery behind them. Let’s get started!

Understanding Quadratic Equations

Quadratic equations are a fundamental aspect of algebra, and they play a crucial role in various fields such as physics, engineering, and economics. These equations involve variables raised to the power of two (hence quadratic) and consist of three terms: the square term, the linear term, and the constant term.

The general form of a quadratic equation can be represented as ax^2 + bx + c = 0. In this specific case, we have the equation 4x^2 – 5x – 12 = 0. The goal is to find the values of x that satisfy this equation.

To solve quadratic equations like this one, there are several methods available. One common approach is factoring. This involves breaking down the equation into factors that can then be set equal to zero.

Another method is using the quadratic formula: (-b ± √(b^2 – 4ac))/(2a). By substituting the coefficients from our given equation into this formula, we can determine the solutions for x.

Completing the square is yet another technique used to solve quadratic equations. This method involves manipulating an equation to create a perfect square trinomial which can then be easily solved.

Understanding these techniques allows us to efficiently tackle any given quadratic equation and find its solutions accurately. Moreover, comprehending concepts related to quadratics such as discriminant and nature of roots helps us gain further insights into their characteristics.

In addition to mathematical applications, understanding quadratics also has real-world relevance. For example, these equations are used in physics calculations involving motion or projectile trajectories. They also come into play when analyzing business profit models or optimizing resource allocation in engineering projects.

Grasping how quadratic equations work opens up a world of possibilities for problem-solving across various disciplines. So let’s dive deeper into solving our specific example – 4x^2 – 5x – 12 = 0 – and explore the different methods available to find its solutions.

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

Solving quadratic equations can seem intimidating at first, but with the right approach, it becomes much more manageable. In this section, we will explore different methods to solve the equation 4x^2 – 5x – 12 = 0.

Step 1: Factoring Method

One of the methods to solve quadratic equations is the factoring method. This approach involves expressing the given equation as a product of two binomials, which can then be set equal to zero and solved for the variables.

To begin with, let’s take a closer look at our example equation: 4x^2 – 5x – 12 = 0. Our goal is to find two expressions that multiply together to give this quadratic equation.

In order to factor this equation, we need to identify two numbers whose product equals the product of the coefficient of x^2 (which is 4) and the constant term (which is -12). In this case, those numbers are -6 and +2.

Next, we rewrite the middle term (-5x) using these two numbers. We split it into -6x and +2x:

4x^2 – 6x + 2x – 12 =0

Now we group pairs containing common factors:

(4x^2-6)x +(2x-12)=0

We can now factor by grouping:


Setting each factor equal to zero gives us our solutions:

(2*x-3)=0 or (+1)=0

Step 2: Quadratic Formula

Now that we understand the basics of quadratic equations, let’s move on to step 2 – solving the equation using the quadratic formula. The quadratic formula is a powerful tool that allows us to find the solutions (or roots) of any quadratic equation.

The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, where a, b, and c are constants:

x = (-b ± √(b^2 – 4ac)) / 2a

It may look a bit intimidating at first glance, but don’t worry! Let’s break it down. The “±” symbol means we will have two possible solutions – one with a plus sign and one with a minus sign. This accounts for both positive and negative values of the square root.

To apply this formula to our example equation 4x^2 – 5x -12 = 0, we can simply substitute the values of a, b, and c into the formula:

x = (-(-5) ± √((-5)^2 – (4)(4)(-12))) / (2)(4)

Simplifying further:

x = (5 ± √(25 +192)) /8
= (5 ± √217)/8

And there you have it – our solutions for x!

Remember, practice makes perfect when it comes to mastering any mathematical concept. Don’t be afraid to work through several examples using different methods until you feel confident in your abilities.

Next up in our journey through quadratic equations: Step 3 – Completing the Square! Stay tuned.

Step 3: Completing the Square

Now that we have explored factoring and the quadratic formula, let’s dive into another method to solve quadratic equations: completing the square. This technique involves manipulating the equation to create a perfect square trinomial, which can then be easily solved.

To begin, we want to rewrite our equation in a specific form: ax^2 + bx = c. In our case, it would be 4x^2 – 5x = 12. To complete the square, we need to add and subtract a constant term that will make this expression a perfect square trinomial.

The first step is to divide all terms by the coefficient of x^2 (a). In this case, it means dividing everything by 4. So now our equation becomes x^2 – (5/4)x = 3.

Next, take half of the coefficient of x (b) and square it [(b/2)^2]. For our equation, half of -5/4 is -5/8; squaring it gives us 25/64.

Now comes the crucial part. Add and subtract that value inside parentheses after (-5/4)x term:

x^2 – (5/4)x + (25/64) – (25/64) = 3

Rearranging this expression gives us:

(x – (5/8))^2 – ((25)/(16))=3

Simplifying further,


This resulting expression can now be written as \(a^{²}\)-c=d where d=0 since we are trying to solve for x when \(ax^(²)+bx+c=0\). Now you can apply your knowledge from Step One or Two!

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