Solving Quadratic Equations: Step-by-Step With The Formula

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Hey math enthusiasts! Let's dive into the world of quadratic equations and unlock the secrets of the quadratic formula. We're going to tackle the equation 5x=6x2−35x = 6x^2 - 3 and find the values of x that make it all work. Trust me, it's not as scary as it sounds, and by the end of this, you'll be a quadratic equation master!

Understanding Quadratic Equations

So, what exactly is a quadratic equation? Well, it's an equation that can be written in the form ax2+bx+c=0ax^2 + bx + c = 0, where a, b, and c are constants, and a is not equal to zero. The highest power of the variable (in this case, x) is 2. The quadratic formula is our go-to tool for solving these types of equations. It's a lifesaver when factoring or other methods get tricky. Think of it as your mathematical Swiss Army knife!

Our equation, 5x=6x2−35x = 6x^2 - 3, needs a little makeover before we can use the formula. We need to rearrange it to match the standard form ax2+bx+c=0ax^2 + bx + c = 0. So, let's bring everything to one side:

6x2−5x−3=06x^2 - 5x - 3 = 0

Now, we can clearly see that a = 6, b = -5, and c = -3. We're ready to put the quadratic formula to work!

The Quadratic Formula: Your Mathematical Superhero

Alright, here comes the star of the show: the quadratic formula! It's a formula that gives you the solutions (also known as roots) of any quadratic equation. The formula is:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Don't worry, it looks a little intimidating at first, but with practice, it becomes second nature. The ±\pm symbol means we'll get two possible solutions – one where we add the square root, and one where we subtract it. Let's plug in the values of a, b, and c from our equation 6x2−5x−3=06x^2 - 5x - 3 = 0.

  • a = 6
  • b = -5
  • c = -3

Substituting these values into the formula, we get:

x=−(−5)±(−5)2−4⋅6⋅(−3)2⋅6x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot (-3)}}{2 \cdot 6}

See? We're already making progress! It's all about carefully substituting the values and then simplifying step by step.

Step-by-Step Calculation: Putting it all Together

Okay, let's break down the calculation. First, simplify the terms inside the square root and the numerator:

x=5±25+7212x = \frac{5 \pm \sqrt{25 + 72}}{12}

Notice how the negative signs work together to become positive! Next, simplify the expression under the square root:

x=5±9712x = \frac{5 \pm \sqrt{97}}{12}

And there we have it! The two solutions for x are:

  • x=5+9712x = \frac{5 + \sqrt{97}}{12}
  • x=5−9712x = \frac{5 - \sqrt{97}}{12}

These are the values of x that satisfy the original equation 5x=6x2−35x = 6x^2 - 3. Pretty cool, right? These are the exact solutions, and you can approximate them using a calculator if you need a decimal value.

Explanation of the Steps

  1. Rearrange the Equation: The first step involved rewriting the equation in the standard quadratic form (ax2+bx+c=0ax^2 + bx + c = 0). This is crucial because the quadratic formula is designed to solve equations in this format. By rearranging 5x=6x2−35x = 6x^2 - 3 to 6x2−5x−3=06x^2 - 5x - 3 = 0, we clearly identified the coefficients a, b, and c, which are essential for applying the quadratic formula.
  2. Identify Coefficients: Once the equation was in the correct form, we identified the coefficients: a = 6, b = -5, and c = -3. These values represent the numerical constants in front of the x2x^2, x, and the constant term, respectively. Accurate identification of these coefficients is critical for the correct application of the formula.
  3. Apply the Quadratic Formula: We then substituted these coefficients into the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. This step is purely mechanical, involving the direct substitution of a, b, and c into the formula. The correct substitution is the foundation for the rest of the solution.
  4. Simplify the Expression: After substituting the values, the expression was simplified step by step. This included simplifying the terms inside the square root (the discriminant) and performing the arithmetic operations. This step requires careful attention to detail, particularly when dealing with negative numbers and exponents, to avoid errors.
  5. Calculate the Solutions: Finally, we calculated the two possible solutions for x using the ±\pm symbol. This resulted in two distinct values for x: x=5+9712x = \frac{5 + \sqrt{97}}{12} and x=5−9712x = \frac{5 - \sqrt{97}}{12}. These two values are the roots of the quadratic equation, representing the points where the equation equals zero.

Decoding the Solutions: What Does it Mean?

So, what do these solutions actually tell us? In the context of a quadratic equation, the solutions (also known as roots or zeros) represent the x-intercepts of the parabola that the equation defines. If you were to graph the equation y=6x2−5x−3y = 6x^2 - 5x - 3, these x-intercepts would be the points where the parabola crosses the x-axis. They are the values of x for which y equals zero.

In our case, the solutions 5+9712\frac{5 + \sqrt{97}}{12} and 5−9712\frac{5 - \sqrt{97}}{12} are the x-coordinates of the points where the parabola intersects the x-axis. These points are key to understanding the behavior of the quadratic function. The solutions give you vital information about the function's shape and position on the graph. For instance, knowing the x-intercepts helps you determine if the parabola opens upwards (positive a) or downwards (negative a), and where the vertex (the minimum or maximum point) is located.

For practical applications, these solutions can represent various real-world scenarios. Imagine the equation models the trajectory of a ball thrown in the air. The solutions would tell you where the ball hits the ground. Or, if the equation models the profit of a business, the solutions would show you the break-even points. The solutions are not just abstract numbers; they offer insight into a problem's characteristics and potential outcomes.

Alternative Methods: Exploring Other Approaches

While the quadratic formula is a universal method for solving quadratic equations, other approaches, such as factoring and completing the square, can be more efficient in certain situations. Let's briefly look at these methods.

Factoring: The Elegant Approach

Factoring involves breaking down the quadratic expression into two simpler expressions, which can be multiplied to give the original expression. For instance, if you could factor 6x2−5x−36x^2 - 5x - 3 into the form (px+q)(rx+s)(px + q)(rx + s), you would set each factor to zero and solve for x. This method works quickly if the quadratic equation can be easily factored, which is not always the case.

To use factoring effectively, you need to find two numbers whose product is equal to ac (in our equation, 6 * -3 = -18) and whose sum is equal to b (-5). Sometimes it can be time-consuming to identify these numbers.

Completing the Square: A Step-by-Step Approach

Completing the square involves transforming the quadratic equation into a perfect square trinomial, which can be easily solved. This method is especially useful when the leading coefficient a is not 1, as it requires a bit more manipulation. The general steps involve moving the constant term to the right side of the equation, dividing by a (if a ≠ 1), adding and subtracting (b/2a)2(b/2a)^2 to both sides, and then factoring the perfect square trinomial. This method, while accurate, can be a bit more complex than the quadratic formula or factoring, especially when the coefficients are not simple integers.

Choosing the Right Method

The choice of method depends on the equation and your preference. The quadratic formula is a reliable choice for any quadratic equation, but factoring can be faster if applicable. Completing the square is a great way to understand the concept and is useful when you can't factor easily. In our example, the quadratic formula was the most straightforward choice because the equation did not factor easily.

Practice Makes Perfect: More Examples

Let's try a few more examples to cement your understanding! Here are a couple of practice problems with solutions:

  1. Solve for x: x2+4x+3=0x^2 + 4x + 3 = 0

    • Using the quadratic formula: a = 1, b = 4, c = 3
    • x=−4±42−4⋅1⋅32⋅1x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}
    • x=−4±16−122x = \frac{-4 \pm \sqrt{16 - 12}}{2}
    • x=−4±42x = \frac{-4 \pm \sqrt{4}}{2}
    • x=−4±22x = \frac{-4 \pm 2}{2}
    • x=−1,−3x = -1, -3

  2. Solve for x: 2x2−7x+3=02x^2 - 7x + 3 = 0

    • Using the quadratic formula: a = 2, b = -7, c = 3
    • x=7±(−7)2−4⋅2⋅32⋅2x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}
    • x=7±49−244x = \frac{7 \pm \sqrt{49 - 24}}{4}
    • x=7±254x = \frac{7 \pm \sqrt{25}}{4}
    • x=7±54x = \frac{7 \pm 5}{4}
    • x=3,12x = 3, \frac{1}{2}

Keep practicing, and you'll become a quadratic equation wizard in no time! Remember to always double-check your work and be mindful of your calculations. The more you practice, the easier it becomes.

Conclusion: You've Got This!

Fantastic job, guys! You've successfully navigated the quadratic formula and solved for x. Remember, the key is to understand the steps and practice regularly. Don't hesitate to revisit the examples and try more problems on your own. Keep up the awesome work, and your math skills will continue to grow!

So, looking back at our original question and the available choices, the correct answer is:

A. 5±9712\frac{5 \pm \sqrt{97}}{12}