**Introduction:**

Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0 Are Polynomial Equations Of Degree 2, Which Are Known As Quadratic Equations.

We Are Given The Quadratic Equation In This Instance:

04x^2 – 5x – 12 = 04×2 – 5x – 12 = 4×2 – 5x – 12 = 0

**Step-By-Step Solution:**

We Have Several Options For Solving The Quadratic Equation: Factoring, Completing The Square, And Applying The Quadratic Formula. Since The Quadratic Formula Is A Dependable Technique That May Be Used To Any Quadratic Problem, We Shall Employ It In This Instance.

**Formula For Quadratic Equations:**

Solving quadratic equations is a cornerstone of algebra and a foundational skill in mathematics. The equation “4x^2 – 5x – 12 = 0” presents a unique challenge and learning opportunity. This article delves deep into the techniques and strategies for solving this equation, ensuring that readers come away with a clear understanding and the ability to tackle similar problems.

**Understanding Quadratic Equations:**

Quadratic equations are polynomials of degree two, typically ax^2 + bx + c = 0, where a, b, and c are constants. The x represents the variable or unknown we are trying to solve. These equations are pivotal in various fields, from physics and engineering to finance and statistics.

A clear example is the quadratic equation in focus, 4x^2 – 5x – 12 = 0. Here, a = 4, b = -5, and c = -12. Understanding the structure of quadratic equations is crucial for solving them, as the values of a, b, and c directly influence the solving strategy and the nature of the solutions.

**Factoring as a Strategy:**

Factoring is one of the primary methods used to solve quadratic equations. It involves rewriting the quadratic in a product of two binomials. However, not all quadratics are easily factorable, especially when dealing with more significant coefficients or when the equation does not simplify neatly. Regarding 4x^2 – 5x – 12 = 0, factoring can be challenging due to the coefficients involved. Nevertheless, one can uncover the factors that solve the equation with some practice and technique.

**The Process of Factoring:**

Factoring aims to find two numbers that multiply to give the product of a and c and add to provide b. For our equation, we look for two numbers that multiply to (4)(-12) = -48 and add to -5. This step requires trial and error but is facilitated by a strong understanding of the number of properties and factors.

Once the appropriate numbers are found, the equation is rewritten in a factored form, which can then be set to zero to solve for the values of x. This method is highly effective for easily factorable equations, providing a quick and straightforward solution.

**The Quadratic Formula: A Universal Solver:**

When factoring is too cumbersome or impossible, the quadratic formula offers a fail-safe solution. This formula, x = [-b ± √(b^2 – 4ac)] / 2a, works for any quadratic equation. It directly utilizes the coefficients a, b, and c to find the solutions of x.

**Applying the Quadratic Formula to 4x^2 – 5x – 12 = 0:**

To solve 4x^2 – 5x – 12 = 0 using the quadratic formula, we substitute the values of a, b, and c into the formula. This process demystifies the equation, turning it into a straightforward calculation. The beauty of the quadratic formula lies in its universality and reliability, ensuring that even the most complex equations can be solved.

**Completing the Square: An Alternate Route:**

Completing the square is another technique for solving quadratic equations. It involves transforming the equation into a perfect square trinomial, from which the value of x can be easily derived. This method is beneficial for understanding the quadratic formula’s derivation and solving equations when other methods are more complex.

**How to Complete the Square:**

Completing the square requires rearranging and manipulating the equation so that the left side becomes a perfect square. The equation 4x^2 – 5x – 12 = 0 involves:

- Dividing by the leading coefficient (if not 1).
- Moving the constant term to the other side.
- Adding a specific value to both sides to create a perfect square trinomial.

This method finds the solution and offers insight into the equation’s structure.

**Graphical Solutions: Visualizing the Equation:**

Graphing is a powerful tool for visualizing and solving quadratic equations. Plotting the equation on a coordinate plane allows one to see the parabola it forms and identify its roots—the points at which it intersects the x-axis.

**Solving 4x^2 – 5x – 12 = 0 Graphically:**

To solve our equation graphically, one would plot y = 4x^2 – 5x – 12 and look for the x-values where y equals zero. This approach provides a visual understanding of the solutions and the behavior of quadratic functions, complementing the algebraic techniques.

**Here Is The Quadratic Formula:**

\Frac{-B \Pm \Sqrt{B^2 – 4ac}} = X=−B±B2−4ac2ax{2a}X=2a−B±B2−4ac}, Where The Coefficients Of The Quadratic Equation Ax2+Bx+C=0ax^2 + Bx + C = 0ax2+Bx+C=0 Are Aaa, Bbb, And Ccc…

The Coefficients For Our Equation 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 Are As Follows:

• A=4a = 4a=4

• B=-5b = -5b = −5

• C=-12c = -12c = -12

**Step 1: Determine The Discriminant:**

The Discriminant Δ\Deltaδ Is Computed As Follows And Is A Component Of The Quadratic Formula Under The Square Root:

Δ=B2−4acdelta Is Equal To B^2 – 4ac.Δ=B2−4ac

Changing The Values Of Bbb, Ccc, And Aaa:

Δ=(−5)2−4(4)(−12)\(-5)^2 – 4(4)(-12) Is Delta.Δ=(−5)2−4(4)(−12) Δ=25+192\Delta Is Equal To 25 Plus 192.Δ=25+192 Δ=217\Delta = 217Δ = 217

We Get Two Different Real Solutions Since The Discriminant Is Positive (Δ>0\Delta > 0Δ>0).

**Step 2: Utilize The Formula For Quadratics:**

Enter Aaa, Bbb, And Δ\Deltaδ Values Into The Quadratic Formula Now:

X=Frac{-(-5) \Pm \Sqrt{217}} = −(−5)±2172(4){2(4)}X=5±2178x = \Frac{5 \Pm \Sqrt{217}}{8}X=85±217; X=2(4)−(−5)±217

Step 3: Make The Solutions Simpler

The Answers Are As Follows:

X2=5−2178x_2 = \Frac{5 – \Sqrt{217}}{8}X2=85−217 X1=5+2178x_1 = \Frac{5 + \Sqrt{217}}{8}X1=85+217

**Checking The Solutions:**

We Can Reintroduce Our Solutions Into The Original Equation And Check That They Fulfill It To Make Sure Our Solutions Are Accurate.

**Confirmation Regarding X1x_1x1:**

The Equation X1=5+2178x_1 = \Frac{5 + \Sqrt{217}}{8}X1=85+217

To Solve 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0, Substitute X1x_1x1 Into 4(5+2178)2−5(5+2178)5\Left(\Frac{5 + \Sqrt{217}}{8}\Right) – −124\Left(\Frac{5 + \Sqrt{217}}{8}\Right)^2 – 124 (85 + 217)2−5(85+217)-12

This Reduces To Zero, Indicating That X1x_1x1 Is A Solution.

**Confirmation Regarding X2x_2x2:**

\Frac{5 – \Sqrt{217}}{8}X2=85−217; X2=5−2178x_2

In 4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0, Substitute X2x_2x2: 4(5−2178)2−5(5−2178)−124\Left(\Frac{5 – \Sqrt{217}}{8}\Right)^2 – 5\Left(\Frac{5 – \Sqrt{217}}{8}\Right)^2 – 5\Left(\Frac{5 – \Sqrt{217}}{8}\Right) – 124(85- 217)2−5(85- 217)−12

This Likewise Reduces To 0, Indicating That X2x_2x2 Is A Solution.

**Summary:**

4×2−5x−12=04x^2 – 5x – 12 = 04×2−5x−12=0 Has The Following Solutions: X1=5+2178x_1 = \Frac{5 + \Sqrt{217}}{8}X1=85+217 X2=5−2178x_2 = \Frac{5 – \Sqrt{217}}{8}X2=85−217

The Quadratic Formula Was Used To Find These Answers, Which Were Then Confirmed By Substituting Them Back Into The Original Problem.