Solve Quadratic Equations With Quadratic Formula

by Sebastian MΓΌller 49 views

Understanding Quadratic Equations

Before diving into the quadratic formula, let's make sure we're all on the same page about what a quadratic equation actually is. A quadratic equation is a polynomial equation of the second degree. That basically means the highest power of the variable (usually x) is 2. The general form of a quadratic equation is:

ax2+bx+c=0ax^2 + bx + c = 0

Where:

  • a, b, and c are coefficients (numbers), and a is not equal to 0 (because if a were 0, it wouldn't be a quadratic equation anymore, would it?).
  • x is the variable we're trying to solve for.

Examples of Quadratic Equations

Here are a few examples to illustrate what quadratic equations look like:

  • x2+3x+2=0x^2 + 3x + 2 = 0

  • 2x2βˆ’5x+1=02x^2 - 5x + 1 = 0

  • x2βˆ’9=0x^2 - 9 = 0

Why Solve Quadratic Equations?

Now, you might be wondering, why bother solving these things anyway? Well, quadratic equations pop up in all sorts of real-world applications, including physics, engineering, economics, and even computer graphics. They can be used to model projectile motion, calculate areas, and optimize processes. Understanding how to solve them is a crucial skill in many fields.

The Quadratic Formula: Your Superpower

Okay, so we know what quadratic equations are. Now, let's talk about how to solve them. There are a few methods, but the quadratic formula is a reliable and versatile tool that works for any quadratic equation. It's like a superpower for math! The quadratic formula is:

x = \frac{-b old{\pm} \sqrt{b^2 - 4ac}}{2a}

Breaking Down the Formula

Let's dissect this formula to make sure we understand what each part means:

  • x: This represents the solutions (also called roots or zeros) of the quadratic equation. A quadratic equation can have up to two solutions.
  • a, b, and c: These are the coefficients from the standard form of the quadratic equation (ax2+bx+c=0ax^2 + bx + c = 0).
  • Β±\pm: This symbol means β€œplus or minus.” It indicates that there are two possible solutions: one where you add the square root part and one where you subtract it.
  • \sqrt{}: This is the square root symbol. We'll need to calculate the square root of the expression inside.
  • b2βˆ’4acb^2 - 4ac: This part is called the discriminant. It tells us about the nature of the solutions (we'll talk more about this later).

How to Use the Quadratic Formula: A Step-by-Step Guide

Now that we know the formula, let's see how to use it to solve a quadratic equation. Here's a step-by-step guide:

  1. Identify a, b, and c: First, rewrite the equation in the standard form (ax2+bx+c=0ax^2 + bx + c = 0) and identify the values of a, b, and c. Make sure you pay attention to the signs (+ or -)!
  2. Plug the values into the formula: Substitute the values of a, b, and c into the quadratic formula.
  3. Simplify: Carefully simplify the expression. Start by calculating the discriminant (b2βˆ’4acb^2 - 4ac).
  4. Calculate the square root: Find the square root of the discriminant. If the discriminant is negative, you'll be dealing with imaginary numbers (more on that later).
  5. Find the two solutions: Use the Β±\pm symbol to calculate the two possible solutions: one with addition and one with subtraction.
  6. Simplify further: If possible, simplify the solutions to their simplest forms.

Example: Solving $x^2 + 3x + 6 = 0$

Let's apply the quadratic formula to solve the equation $x^2 + 3x + 6 = 0$. This is the equation in your original question, so we'll go through it step-by-step.

  1. Identify a, b, and c:

    In this equation:

    • a = 1 (the coefficient of x2x^2)
    • b = 3 (the coefficient of x)
    • c = 6 (the constant term)

  2. Plug the values into the formula:

    Substitute these values into the quadratic formula:

    x=βˆ’3Β±32βˆ’4(1)(6)2(1)x = \frac{-3 \pm \sqrt{3^2 - 4(1)(6)}}{2(1)}

  3. Simplify:

    Let's simplify the expression step-by-step:

    • Calculate the discriminant: 32βˆ’4(1)(6)=9βˆ’24=βˆ’153^2 - 4(1)(6) = 9 - 24 = -15

    • So, our equation now looks like:

      x=βˆ’3Β±βˆ’152x = \frac{-3 \pm \sqrt{-15}}{2}

  4. Calculate the square root: Here's where things get interesting! We have the square root of a negative number (-15). This means our solutions will involve imaginary numbers. Remember that the square root of -1 is defined as i (the imaginary unit).

    • We can rewrite βˆ’15\sqrt{-15} as 15βˆ—βˆ’1=15βˆ—βˆ’1=i15\sqrt{15 * -1} = \sqrt{15} * \sqrt{-1} = i\sqrt{15}

    • Our equation now becomes:

      x=βˆ’3Β±i152x = \frac{-3 \pm i\sqrt{15}}{2}

  5. Find the two solutions: We have two solutions, one with addition and one with subtraction:

    • x1=βˆ’3+i152x_1 = \frac{-3 + i\sqrt{15}}{2}

    • x2=βˆ’3βˆ’i152x_2 = \frac{-3 - i\sqrt{15}}{2}

  6. Simplify further: The solutions are already in their simplest form.

Therefore, the solutions to the quadratic equation $x^2 + 3x + 6 = 0$ are $\frac{-3 + i\sqrt{15}}{2}$ and $\frac{-3 - i\sqrt{15}}{2}$. Guys, that's it! We solved it using the quadratic formula!

Understanding the Discriminant: A Deeper Dive

Remember that part of the quadratic formula we called the discriminant (b2βˆ’4acb^2 - 4ac)? Well, it's more than just a step in the process. The discriminant actually tells us about the nature of the solutions to the quadratic equation. It's like a secret code that reveals what kind of answers we'll get.

Types of Solutions Based on the Discriminant

Here's how the discriminant helps us understand the solutions:

  • If b2βˆ’4ac>0b^2 - 4ac > 0 (positive): The equation has two distinct real solutions. This means the parabola represented by the quadratic equation intersects the x-axis at two different points.
  • If b2βˆ’4ac=0b^2 - 4ac = 0: The equation has one real solution (a repeated root). This means the parabola touches the x-axis at exactly one point (the vertex of the parabola).
  • If b2βˆ’4ac<0b^2 - 4ac < 0 (negative): The equation has two complex (imaginary) solutions. This means the parabola does not intersect the x-axis.

Example: Using the Discriminant

Let's look at a few examples to see how the discriminant works:

  • Equation: x2+2xβˆ’3=0x^2 + 2x - 3 = 0
    • Discriminant: 22βˆ’4(1)(βˆ’3)=4+12=162^2 - 4(1)(-3) = 4 + 12 = 16 (positive)
    • Solutions: Two distinct real solutions
  • Equation: x2+4x+4=0x^2 + 4x + 4 = 0
    • Discriminant: 42βˆ’4(1)(4)=16βˆ’16=04^2 - 4(1)(4) = 16 - 16 = 0
    • Solutions: One real solution (repeated root)
  • Equation: x2+x+1=0x^2 + x + 1 = 0
    • Discriminant: 12βˆ’4(1)(1)=1βˆ’4=βˆ’31^2 - 4(1)(1) = 1 - 4 = -3 (negative)
    • Solutions: Two complex (imaginary) solutions

Tips and Tricks for Mastering the Quadratic Formula

Okay, guys, we've covered a lot! Here are some tips and tricks to help you master the quadratic formula and become a quadratic equation-solving pro:

  • Practice, practice, practice!: The more you practice, the more comfortable you'll become with the formula and the process. Work through as many examples as you can get your hands on.
  • Double-check your signs: One of the most common mistakes is messing up the signs of a, b, or c. Pay close attention to those pluses and minuses!
  • Simplify carefully: Take your time when simplifying the expression, especially when dealing with square roots and fractions. A small mistake can throw off the whole answer.
  • Use the discriminant to predict the solutions: Before you even start plugging numbers into the formula, calculate the discriminant. This will give you a heads-up about what kind of solutions to expect.
  • Don't be afraid of complex numbers: If you get a negative discriminant, don't panic! Just remember the definition of i and keep going. Complex solutions are perfectly valid.
  • Check your answers: Once you've found the solutions, plug them back into the original equation to make sure they work. This is a great way to catch any mistakes.

Conclusion

Guys, you've made it to the end! You now have a solid understanding of how to solve quadratic equations using the quadratic formula. Remember, the key is to understand the formula, practice consistently, and pay attention to detail. With a little effort, you'll be solving quadratic equations like a boss! So, go forth and conquer those equations!

The solution to the example quadratic equation $x^2 + 3x + 6 = 0$ is (D) $\frac{-3 \pm i \sqrt{15}}{2}$.