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

A quadratic equation is a second-degree polynomial equation in a single variable, usually represented as “ax^2 + bx + c = 0,” where ‘a,’ ‘b,’ and ‘c’ are constants, and ‘x’ is the variable we need to find.

To solve quadratic equations, we often use the quadratic formula, which is a powerful tool for finding the values of ‘x.’ The quadratic formula is:

x = (-b Â± âˆš(b^2 – 4ac)) / (2a)

Now, let’s apply this formula to the equation “4x^2 – 5x – 12 = 0.”

Step 1: Identify the Coefficientsof 4x^2 – 5x – 12 = 0

In our equation, “4x^2 – 5x – 12 = 0,” we have ‘a,’ ‘b,’ and ‘c’ as follows:

• a = 4
• b = -5
• c = -12

Step 2: Plug into the Quadratic Formula

Now that we have our coefficients, let’s plug them into the quadratic formula:

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

Step 3: Simplify

Let’s simplify the equation:

x = (5 Â± âˆš(25 + 192)) / 8

x = (5 Â± âˆš217) / 8

Step 4: Find the Two Solutions

Now, we have two possible solutions:

1. x = (5 + âˆš217) / 8, Or
2. x = (5 – âˆš217) / 8

Conclusion

We have successfully solved the quadratic equation “4x^2 – 5x – 12 = 0” using the quadratic formula. The two solutions, provide the values of ‘x’ that satisfy the equation. Quadratic equations are fundamental in mathematics and have various real-world applications, making it essential to understand their solutions.

FAQs:

What is the importance of solving quadratic equations?

Quadratic equations are used in various fields, including physics, engineering, and economics, making their solutions valuable for real-world problem-solving.

Can quadratic equations have no real solutions?

Yes, if the discriminant (the value inside the square root in the quadratic formula) is negative, the equation has no real solutions.

Are there alternative methods to solve quadratic equations?

Yes, besides the quadratic formula, you can also solve quadratic equations by factoring, completing the square, or using graphing methods.

How do I know which solution to use in practical applications?

The choice of solution depends on the context of the problem. Both solutions are valid, so consider the specific problem to determine which one is relevant.

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