The quadratic equation, a cornerstone in algebra, often appears intimidating to students and enthusiasts alike. The equation 4x^2 – 5x – 12 = 0 is a quintessential example that encompasses the fundamental principles of solving quadratic equations. This article delves deep into understanding and solving this equation, offering expert techniques to demystify the process and build confidence in tackling similar problems.

## Understanding the Quadratic Equation

Quadratic equations take the general form ax^2 + bx + c = 0, where ‘a,’ ‘b,’ and ‘c’ are constants. In the equation 4x^2 – 5x – 12 = 0, we identify ‘a’ as 4, ‘b’ as -5, and ‘c’ as -12. The goal is to find the values of ‘x’ that satisfy the equation, known as the roots of the equation.

Quadratic equations can have two real roots, one real root, or two complex roots, depending on the discriminant (b^2 – 4ac). The nature of the origins provides insights into the equation’s solutions, guiding us in choosing the appropriate method to solve it.

## The Discriminant and Its Significance

Before diving into solving 4x ^ 2 – 5x – 12 = 0, it’s crucial to understand the discriminant, which is calculated as D = b^2 – 4ac. The discriminant determines the nature of the roots:

- If D > 0, the equation has two distinct real roots.
- If D = 0, the equation has one real root (a repeated root).
- If D < 0, the equation has two complex roots.

For our equation, the discriminant is calculated as follows: π·=(β5)2β4(4)(β12)=25+192=217*D*=(β5)2β4(4)(β12)=25+192=217

Since the discriminant is positive (217 > 0), the equation 4x^2 – 5x – 12 = 0 has two distinct real roots. This knowledge reassures us that our solutions will be real numbers, paving the way for solving the equation using various techniques.

## Factoring the Quadratic Equation

Factoring is a method that involves expressing the quadratic equation as a product of two binomials. This method works efficiently when the equation is easily factorable. However, not all quadratic equations can be factored readily, and additional techniques might be required.

For 4x^2 – 5x – 12 = 0, we seek two numbers that multiply to (a*c) = 4*(-12) = -48 and add to ‘b’ = -5. These numbers are 3 and -16. Rewriting the middle term using these numbers, we get: 4π₯2+3π₯β16π₯β12=04*x*2+3*x*β16*x*β12=0

Next, we group the terms and factors by grouping: (4π₯2+3π₯)+(β16π₯β12)=0(4*x*2+3*x*)+(β16*x*β12)=0 π₯(4π₯+3)β4(4π₯+3)=0*x*(4*x*+3)β4(4*x*+3)=0

Factoring out the common binomial (4x + 3), we obtain: (4π₯+3)(π₯β4)=0(4*x*+3)(*x*β4)=0

Setting each factor to zero gives the solutions: 4π₯+3=0βπ₯=β344*x*+3=0β*x*=β43β π₯β4=0βπ₯=4*x*β4=0β*x*=4

Thus, the roots of the equation are π₯=β34*x*=β43β and π₯=4*x*=4.

## The Quadratic Formula

When factoring is not straightforward, the quadratic formula offers a reliable method to solve any quadratic equation. The quadratic formula is given by: π₯=βπΒ±π2β4ππ2π*x*=2*a*β*b*Β±*b*2β4*a**c*ββ

Applying this formula to 4x^2 – 5x – 12 = 0, we substitute ‘a’ = 4, ‘b’ = -5, and ‘c’ = -12: π₯=β(β5)Β±(β5)2β4(4)(β12)2(4)*x*=2(4)β(β5)Β±(β5)2β4(4)(β12)ββ π₯=5Β±25+1928*x*=85Β±25+192ββ π₯=5Β±2178*x*=85Β±217ββ

This results in two distinct roots: π₯=5+2178*x*=85+217ββ π₯=5β2178*x*=85β217ββ

These roots are the solutions to the equation, and their exact values can be approximated using a calculator for practical applications.

## Completing the Square

Completing the square is another method for solving quadratic equations. This method is handy for understanding how the equation transforms into vertex form. The process involves creating a perfect square trinomial from the quadratic equation.

Starting with 4x^2 – 5x – 12 = 0, we first divide by ‘a’ (4) to normalize the coefficient of x^2: π₯2β54π₯β3=0*x*2β45β*x*β3=0

Next, we isolate the constant term: π₯2β54π₯=3*x*2β45β*x*=3

To complete the square, we add and subtract (π2π)2(2*a**b*β)2: π₯2β54π₯+(58)2=3+(58)2*x*2β45β*x*+(85β)2=3+(85β)2 π₯2β54π₯+2564=3+2564*x*2β45β*x*+6425β=3+6425β (π₯β58)2=21764(*x*β85β)2=64217β

Taking the square root of both sides, we obtain: π₯β58=Β±21764*x*β85β=Β±64217ββ π₯=58Β±2178*x*=85βΒ±8217ββ

This simplifies to π₯=5+2178*x*=85+217ββ π₯=5β2178*x*=85β217ββ

These solutions match those obtained using the quadratic formula, demonstrating the consistency and reliability of different methods in solving quadratic equations.

## Graphical Interpretation

Visualizing the quadratic equation on a graph provides a deeper understanding of the solutions. The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of ‘a.’ For 4x^2 – 5x – 12 = 0, ‘a’ is positive (4), so the parabola opens upwards.

The roots of the equation, π₯=β34*x*=β43β and π₯=4*x*=4, correspond to the x-intercepts of the parabola. The vertex of the parabola, given by the formula π₯=βπ2π*x*=β2*a**b*β, provides the axis of symmetry. For our equation: π₯=ββ52(4)=58*x*=β2(4)β5β=85β

Substituting π₯=58*x*=85β into the equation to find the y-coordinate of the vertex: π¦=4(58)2β5(58)β12*y*=4(85β)2β5(85β)β12 π¦=4(2564)β258β12*y*=4(6425β)β825ββ12 π¦=10064β20064β76864*y*=64100ββ64200ββ64768β π¦=100β200β76864*y*=64100β200β768β π¦=β86864=β13.5625*y*=64β868β=β13.5625

The vertex of the parabola is (58,β13.5625)(85β,β13.5625), indicating the lowest point of the graph. Understanding the graphical interpretation helps visualize the quadratic equation’s behavior and confirm the solutions’ accuracy.

## Conclusion

Solving the quadratic equation 4x^2 – 5x – 12 = 0 involves understanding various techniques, each offering a unique approach to finding the roots. From factoring and the quadratic formula to completing the square and graphical interpretation, mastering these methods equips us with the tools to tackle any quadratic equation confidently.

Whether you’re a student grappling with algebra or an enthusiast seeking to deepen your mathematical prowess, these expert techniques provide a comprehensive guide to demystifying quadratic equations. Embrace these strategies, and you’ll find that solving equations like 4x^2 – 5x – 12 = 0 becomes a manageable and rewarding challenge.