Solve 2x^2 - 24x + 54 = 0: Find X Values!

by Henrik Larsen 42 views

Hey guys! Today, let's dive into the fascinating world of quadratic equations and learn how to solve them. Specifically, we're going to tackle the equation 2x2−24x+54=0{2x^2 - 24x + 54 = 0} and find the values of x{x} that make this equation true. Don't worry if you're feeling a bit rusty on your algebra – we'll break it down step by step so everyone can follow along. Get ready to unleash your inner mathlete!

Understanding Quadratic Equations

Before we jump into solving our specific equation, let's take a moment to understand what quadratic equations are all about. A quadratic equation is a polynomial equation of the second degree. This means the highest power of the variable (in our case, x{x}) is 2. The general form of a quadratic equation is:

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

Where a{a}, b{b}, and c{c} are constants, and a{a} is not equal to 0 (otherwise, it would be a linear equation). In our equation, 2x2−24x+54=0{2x^2 - 24x + 54 = 0}, we can identify the coefficients as follows:

  • a=2{a = 2}
  • b=−24{b = -24}
  • c=54{c = 54}

Now that we know what a quadratic equation is, let's explore the methods we can use to solve them. There are several techniques, but we'll focus on factoring and using the quadratic formula.

Methods for Solving Quadratic Equations

There are primarily three methods to solve quadratic equations:

  1. Factoring: This method involves expressing the quadratic equation as a product of two linear factors. It's a straightforward approach when the equation can be factored easily.
  2. Completing the Square: This technique involves manipulating the equation to form a perfect square trinomial, which can then be easily solved.
  3. Quadratic Formula: The quadratic formula is a universal method that works for all quadratic equations, regardless of whether they can be factored easily. It's a powerful tool, and we'll be using it today.

In our case, we'll start by trying to factor the equation, but if that proves difficult, we'll confidently turn to the quadratic formula. It's like having a trusty Swiss Army knife in our mathematical toolkit!

Solving the Equation 2x2−24x+54=0{2x^2 - 24x + 54 = 0}

Okay, let's get down to business and solve our equation: 2x2−24x+54=0{2x^2 - 24x + 54 = 0}. The first thing we should always do when tackling a quadratic equation is to see if we can simplify it. Notice that all the coefficients (2, -24, and 54) are divisible by 2. This means we can divide the entire equation by 2 to make our lives easier: 2x22−24x2+542=02{\frac{2x^2}{2} - \frac{24x}{2} + \frac{54}{2} = \frac{0}{2}}

This simplifies to: x2−12x+27=0{x^2 - 12x + 27 = 0}

Ah, much better! Now we have a simpler equation to work with. Let's try factoring this quadratic expression.

Factoring the Quadratic Expression

Factoring involves finding two binomials that, when multiplied together, give us our quadratic expression. We're looking for two numbers that:

  • Multiply to give us the constant term (27)
  • Add up to give us the coefficient of the x{x} term (-12)

Think about the factors of 27. We have 1 and 27, and 3 and 9. Which pair can we use to get -12 when we add them? Well, if we use -3 and -9:

  • (-3) * (-9) = 27
  • (-3) + (-9) = -12

Bingo! So, we can factor our quadratic expression as follows: (x−3)(x−9)=0{(x - 3)(x - 9) = 0}

Finding the Values of x{x}

Now that we've factored the equation, we can use the zero-product property. This property states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if A∗B=0{A * B = 0}, then either A=0{A = 0} or B=0{B = 0} (or both).

Applying this to our factored equation, we have:

(x−3)(x−9)=0{(x - 3)(x - 9) = 0}

So, either x−3=0{x - 3 = 0} or x−9=0{x - 9 = 0}. Let's solve each of these equations:

  1. If x−3=0{x - 3 = 0}, then adding 3 to both sides gives us x=3{x = 3}.
  2. If x−9=0{x - 9 = 0}, then adding 9 to both sides gives us x=9{x = 9}.

Therefore, the solutions to our quadratic equation are x=3{x = 3} and x=9{x = 9}. We've cracked the code!

The Solutions: Putting it All Together

We've successfully solved the quadratic equation 2x2−24x+54=0{2x^2 - 24x + 54 = 0} by factoring. We found that the values of x{x} that satisfy the equation are 3 and 9. Now, let's present our solutions in the requested format, from smallest to largest:

menor x=3{\text{menor } x = 3} maior x=9{\text{maior } x = 9}

So, the correct answer from your provided alternatives is:

  • A) x1=3{x_1 = 3} e x2=9{x_2 = 9}

Awesome! We've not only found the solutions but also understood the process behind solving quadratic equations by factoring. Pat yourselves on the back, guys!

Mastering Quadratic Equations: Further Exploration

Now that you've conquered this quadratic equation, you're well on your way to mastering this important algebraic concept. But the journey doesn't end here! There's always more to learn and explore. Let's briefly touch upon some additional aspects of quadratic equations.

The Quadratic Formula: A Universal Solution

As we mentioned earlier, the quadratic formula is a powerful tool that can solve any quadratic equation, even those that are difficult or impossible to factor. The formula is:

x=−b±b2−4ac2a{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}

Where a{a}, b{b}, and c{c} are the coefficients of the quadratic equation in the standard form ax2+bx+c=0{ax^2 + bx + c = 0}. The ±{\pm} symbol means we have two possible solutions, one with addition and one with subtraction.

Let's see how the quadratic formula would work for our equation 2x2−24x+54=0{2x^2 - 24x + 54 = 0} (or the simplified version x2−12x+27=0{x^2 - 12x + 27 = 0}).

For x2−12x+27=0{x^2 - 12x + 27 = 0}, we have:

  • a=1{a = 1}
  • b=−12{b = -12}
  • c=27{c = 27}

Plugging these values into the quadratic formula:

x=−(−12)±(−12)2−4(1)(27)2(1){x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(27)}}{2(1)}}

x=12±144−1082{x = \frac{12 \pm \sqrt{144 - 108}}{2}}

x=12±362{x = \frac{12 \pm \sqrt{36}}{2}}

x=12±62{x = \frac{12 \pm 6}{2}}

Now we have two possibilities:

  1. x=12+62=182=9{x = \frac{12 + 6}{2} = \frac{18}{2} = 9}
  2. x=12−62=62=3{x = \frac{12 - 6}{2} = \frac{6}{2} = 3}

As you can see, we arrive at the same solutions (3 and 9) using the quadratic formula. This demonstrates the formula's versatility and reliability.

The Discriminant: Unveiling the Nature of Solutions

Within the quadratic formula, the expression b2−4ac{b^2 - 4ac} is called the discriminant. The discriminant tells us about the nature of the solutions to the quadratic equation:

  • If b2−4ac>0{b^2 - 4ac > 0}, the equation has two distinct real solutions (like our example).
  • If b2−4ac=0{b^2 - 4ac = 0}, the equation has one real solution (a repeated root).
  • If b2−4ac<0{b^2 - 4ac < 0}, the equation has two complex solutions (involving imaginary numbers).

In our example, the discriminant is 36, which is greater than 0, confirming that we have two distinct real solutions.

Graphing Quadratic Equations: Parabolas

Quadratic equations have a beautiful graphical representation: they form parabolas. A parabola is a U-shaped curve, and the solutions to the quadratic equation correspond to the points where the parabola intersects the x-axis (also known as the roots or zeros of the equation).

The parabola for our equation x2−12x+27=0{x^2 - 12x + 27 = 0} would intersect the x-axis at x=3{x = 3} and x=9{x = 9}. Understanding the graphical representation can provide valuable insights into the behavior of quadratic equations.

Conclusion: Keep Exploring the World of Math!

Congratulations on mastering the solution to this quadratic equation! We've explored the concepts of factoring, the quadratic formula, the discriminant, and the graphical representation of quadratic equations. Remember, guys, practice makes perfect. The more you work with these concepts, the more comfortable and confident you'll become.

So, keep exploring the fascinating world of math, and don't be afraid to tackle new challenges. You've got this!