Solving Quadratic Equations Finding Real Solutions For 2y^2 + 9y + 3 = 4

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Hey guys! Today, we're diving into the world of quadratic equations. We're going to tackle the equation 2y² + 9y + 3 = 4 and find all its real solutions in the simplest form. Don't worry, it might seem intimidating at first, but we'll break it down step by step. So, grab your pencils and let's get started!

Understanding Quadratic Equations

Before we jump into solving, let's quickly recap what quadratic equations are all about. A quadratic equation is a polynomial equation of the second degree. The general form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants, and a is not equal to zero. The solutions to a quadratic equation are also known as its roots or zeros, and they represent the values of x (or in our case, y) that make the equation true.

Key Components of a Quadratic Equation

  • a: The coefficient of the quadratic term (the term with y²).
  • b: The coefficient of the linear term (the term with y).
  • c: The constant term.

In our equation, 2y² + 9y + 3 = 4, we first need to bring it into the standard form by subtracting 4 from both sides. This gives us 2y² + 9y - 1 = 0. Now, we can clearly see that a = 2, b = 9, and c = -1.

Why Solve Quadratic Equations?

You might be wondering, why bother solving these equations? Well, quadratic equations pop up in many real-world applications, from physics and engineering to economics and computer science. They help model parabolic trajectories, optimize areas, and analyze growth rates, among other things. So, understanding how to solve them is a valuable skill!

Methods for Solving Quadratic Equations

There are several methods we can use to solve quadratic equations, each with its own advantages and disadvantages. The most common methods include:

  1. Factoring: This method involves breaking down the quadratic expression into the product of two binomials. It's quick and easy when it works, but it's not always applicable.
  2. Completing the Square: This method involves manipulating the equation to create a perfect square trinomial. It's a bit more involved than factoring, but it works for any quadratic equation.
  3. Quadratic Formula: This is a universal formula that provides the solutions to any quadratic equation. It might seem intimidating at first, but it's a reliable method.

For our equation, 2y² + 9y - 1 = 0, factoring isn't straightforward, so we'll use the quadratic formula. It's like having a Swiss Army knife for quadratic equations – it always gets the job done!

Applying the Quadratic Formula

The quadratic formula is given by:

y = (-b ± √(b² - 4ac)) / (2a)

Where a, b, and c are the coefficients from our quadratic equation ax² + bx + c = 0. Remember, in our case, a = 2, b = 9, and c = -1.

Plugging in the Values

Let's plug these values into the quadratic formula:

y = (-9 ± √(9² - 4 * 2 * (-1))) / (2 * 2)

Simplifying the Expression

Now, we need to simplify the expression step by step:

  1. Calculate the square: 9² = 81
  2. Calculate the product inside the square root: 4 * 2 * (-1) = -8
  3. Subtract the product from 81: 81 - (-8) = 81 + 8 = 89
  4. Multiply in the denominator: 2 * 2 = 4

So, our equation becomes:

y = (-9 ± √89) / 4

The Two Solutions

The ± sign in the formula means we actually have two solutions:

  1. y₁ = (-9 + √89) / 4
  2. y₂ = (-9 - √89) / 4

These are the two real solutions to our equation in their simplest form. We can't simplify the square root of 89 any further because 89 is a prime number. So, we're done!

Checking Our Solutions

To make sure we haven't made any mistakes, it's always a good idea to check our solutions. We can do this by plugging each solution back into the original equation 2y² + 9y + 3 = 4 and verifying that it holds true.

Checking y₁ = (-9 + √89) / 4

Let's substitute y₁ into the equation:

2((-9 + √89) / 4)² + 9((-9 + √89) / 4) + 3

This looks a bit complicated, but let's break it down:

  1. Square the term: ((-9 + √89) / 4)² = (81 - 18√89 + 89) / 16 = (170 - 18√89) / 16
  2. Multiply by 2: 2 * ((170 - 18√89) / 16) = (170 - 18√89) / 8
  3. Multiply 9 by the term: 9 * ((-9 + √89) / 4) = (-81 + 9√89) / 4
  4. Now, add all the terms together: (170 - 18√89) / 8 + (-81 + 9√89) / 4 + 3

To add these fractions, we need a common denominator, which is 8. So, we rewrite the equation as:

(170 - 18√89) / 8 + (-162 + 18√89) / 8 + 24 / 8

Now, we can add the numerators:

(170 - 18√89 - 162 + 18√89 + 24) / 8

Simplify: (170 - 162 + 24) / 8 = 32 / 8 = 4

So, the equation holds true for y₁.

Checking y₂ = (-9 - √89) / 4

Now, let's substitute y₂ into the equation:

2((-9 - √89) / 4)² + 9((-9 - √89) / 4) + 3

Following similar steps as above:

  1. Square the term: ((-9 - √89) / 4)² = (81 + 18√89 + 89) / 16 = (170 + 18√89) / 16
  2. Multiply by 2: 2 * ((170 + 18√89) / 16) = (170 + 18√89) / 8
  3. Multiply 9 by the term: 9 * ((-9 - √89) / 4) = (-81 - 9√89) / 4
  4. Now, add all the terms together: (170 + 18√89) / 8 + (-81 - 9√89) / 4 + 3

Again, we need a common denominator, which is 8. So, we rewrite the equation as:

(170 + 18√89) / 8 + (-162 - 18√89) / 8 + 24 / 8

Now, we can add the numerators:

(170 + 18√89 - 162 - 18√89 + 24) / 8

Simplify: (170 - 162 + 24) / 8 = 32 / 8 = 4

So, the equation also holds true for y₂. This confirms that our solutions are correct!

Real-World Applications of Quadratic Equations

As we mentioned earlier, quadratic equations have numerous applications in the real world. Let's explore a few examples:

Physics

In physics, quadratic equations are used to model projectile motion. For instance, the height of a ball thrown into the air can be described by a quadratic equation. Solving the equation helps determine the maximum height the ball reaches and the time it takes to hit the ground.

Engineering

Engineers use quadratic equations to design bridges, buildings, and other structures. For example, the shape of a parabolic arch in a bridge can be modeled using a quadratic equation, ensuring the structure's stability and load-bearing capacity.

Economics

In economics, quadratic equations can be used to model cost, revenue, and profit functions. Businesses can use these models to determine the optimal production level or pricing strategy to maximize their profits.

Computer Science

Quadratic equations are also used in computer science, particularly in algorithms and optimization problems. For example, they can help find the best solution for a given problem by minimizing a quadratic function.

Tips for Solving Quadratic Equations

Solving quadratic equations can become second nature with practice. Here are a few tips to help you along the way:

  1. Always bring the equation to the standard form (ax² + bx + c = 0) before attempting to solve it. This makes it easier to identify the coefficients a, b, and c.
  2. Try factoring first. If the quadratic expression can be easily factored, this is often the quickest method.
  3. If factoring doesn't work, use the quadratic formula. It's a reliable method that works for any quadratic equation.
  4. Simplify your solutions as much as possible. This may involve simplifying square roots or reducing fractions.
  5. Check your solutions by plugging them back into the original equation. This helps catch any mistakes you may have made.

Conclusion

So, guys, we've successfully solved the quadratic equation 2y² + 9y + 3 = 4 and found its real solutions: y₁ = (-9 + √89) / 4 and y₂ = (-9 - √89) / 4. We've also explored the importance of quadratic equations and their applications in various fields. Remember, practice makes perfect, so keep solving those equations! You've got this!