Solving $4r^4 + 12r^3 + 2r^2 = 0$: Best Methods

by ADMIN 50 views

Hey guys! Let's dive into solving the equation $4r^4 + 12r^3 + 2r^2 = 0$. This looks like a fun math puzzle, and we've got a few different ways we could tackle it. We'll explore the most efficient method to find the values of $r$ that make this equation true. Understanding the best approach can save us time and prevent unnecessary headaches. So, let's get started and break down the equation step-by-step!

Understanding the Equation

Before we jump into solving, let's take a good look at our equation: $4r^4 + 12r^3 + 2r^2 = 0$. Notice that each term has $r$ raised to a power, and all terms are set to zero. This hints that factoring might be our best bet. Factoring helps simplify the equation by breaking it down into smaller, more manageable parts. Essentially, we want to rewrite the equation as a product of simpler expressions. This is a common strategy for solving polynomial equations, and it's often the most straightforward method when possible. Spotting common factors early on is key to efficient problem-solving in algebra. So, with this in mind, let's explore the options we have for factoring this particular equation.

Initial Observations

When tackling an equation like this, the first thing we should look for is a Greatest Common Factor (GCF). In our case, the GCF is the largest expression that divides evenly into each term. Looking at the coefficients (4, 12, and 2), we see that 2 is a common factor. Also, each term contains at least $r^2$, making $r^2$ another part of our GCF. So, the GCF for the entire equation is $2r^2$. Recognizing this GCF is crucial because it allows us to simplify the equation significantly right from the start. Factoring out the GCF is like peeling away the outer layers to reveal the core of the problem, making it much easier to handle. Now that we've identified the GCF, let's see how factoring it out changes the equation.

Factoring out the GCF

Let's factor out the GCF, which we've established is $2r^2$, from the equation $4r^4 + 12r^3 + 2r^2 = 0$. When we factor out $2r^2$, we divide each term by $2r^2$ and rewrite the equation. Here's how it looks:

2r2(2r2+6r+1)=02r^2(2r^2 + 6r + 1) = 0

Now, the equation is in a more simplified form. We have $2r^2$ multiplied by a quadratic expression $(2r^2 + 6r + 1)$. This is a significant step forward because it allows us to address each part separately. We've essentially broken down a fourth-degree polynomial into a simpler form by factoring out the common factor. This not only makes the equation easier to solve but also gives us valuable insights into the possible solutions. The next step is to deal with the quadratic expression, which will help us find the remaining roots of the equation. Let's see how we can do that!

Analyzing the Options

Now that we've factored out the GCF, let's consider the options provided for solving the equation $4r^4 + 12r^3 + 2r^2 = 0$:

  • A. Use backwards FOIL and then complete the square.
  • B. Factor by GCF and then use the quadratic formula.
  • C. Factor by GCF and then use backwards FOIL.
  • D. Use the... (The option is incomplete, but we can still analyze the given information.)

Evaluating Option A: Backwards FOIL and Completing the Square

The term "backwards FOIL" typically refers to factoring a quadratic expression. However, after factoring out the GCF, we are left with the quadratic $2r^2 + 6r + 1$. This quadratic doesn't factor easily using simple integer coefficients, which makes backwards FOIL less practical in this scenario. Completing the square is a valid method for solving quadratic equations, but it can be more cumbersome than other methods, especially when the coefficient of the $r^2$ term isn't 1. So, while option A isn't incorrect, it might not be the most efficient path for this particular problem.

Evaluating Option B: Factoring by GCF and Using the Quadratic Formula

This option aligns perfectly with our initial steps. We've already factored out the GCF, simplifying the original equation. Now, we're left with the quadratic expression $2r^2 + 6r + 1$. Since this quadratic doesn't factor easily, the quadratic formula is an excellent choice. The quadratic formula provides a direct method for finding the roots of any quadratic equation in the form $ax^2 + bx + c = 0$. It's a reliable and widely used technique, making option B a strong contender for the best approach.

Evaluating Option C: Factoring by GCF and Using Backwards FOIL

As we discussed in option A, "backwards FOIL" isn't the most effective method for the quadratic $2r^2 + 6r + 1$ because it doesn't factor neatly. While factoring by GCF was a crucial first step, attempting backwards FOIL here would likely lead to frustration and wouldn't efficiently solve the equation. Therefore, option C is less suitable compared to option B.

Initial Conclusion

Based on our analysis, option B seems like the most promising strategy. Factoring by GCF simplified the equation, and the quadratic formula is a reliable method for solving the remaining quadratic expression. Let's proceed with option B and see how it plays out.

Applying the Quadratic Formula

Now, let's use the quadratic formula to solve the quadratic expression $2r^2 + 6r + 1 = 0$. The quadratic formula is given by:

r = rac{-b o rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}}}{ rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}} rac{ ext{2024 年 8 月 17 日}}{ ext{2024 年 8 月 17 日}}}

In our quadratic, $a = 2$, $b = 6$, and $c = 1$. Plugging these values into the formula, we get:

r = rac{-6 ext{2024 年 8 月 17 日} ext{2024 年 8 月 17 日} ext{2024 年 8 月 17 日}}{4}

Simplifying the Solutions

Now, let's simplify the expression under the square root:

36−4(2)(1)=36−8=2836 - 4(2)(1) = 36 - 8 = 28

So, we have:

r = rac{-6 ext{2024 年 8 月 17 日} ext{2024 年 8 月 17 日}}{4}

We can simplify the square root of 28 by factoring out the largest perfect square, which is 4:

ext2024年8月17日=ext2024年8月17日 ext{2024 年 8 月 17 日} = ext{2024 年 8 月 17 日}

Substituting this back into our equation gives:

r = rac{-6 ext{2024 年 8 月 17 日} ext{2024 年 8 月 17 日}}{4}

Now, we can divide both the numerator and the denominator by 2 to simplify further:

r = rac{-3 ext{2024 年 8 月 17 日} ext{2024 年 8 月 17 日}}{2}

Thus, we have two solutions for $r$ from the quadratic formula:

r_1 = rac{-3 + ext{2024 年 8 月 17 日}}{2}

r_2 = rac{-3 - ext{2024 年 8 月 17 日}}{2}

These are the solutions derived from the quadratic portion of the equation. But remember, we factored out a GCF earlier, which also gives us solutions.

Accounting for the GCF

Earlier, we factored out $2r^2$ from the original equation. This means $2r^2 = 0$ is also part of the solution. Solving for $r$ in this case:

2r2=02r^2 = 0

r2=0r^2 = 0

r=0r = 0

Since $r^2 = 0$, this root has a multiplicity of 2, meaning it appears twice as a solution. So, we have $r = 0$ as a repeated root.

Final Solutions

Combining the solutions from the quadratic formula and the GCF, we have the following solutions for the equation $4r^4 + 12r^3 + 2r^2 = 0$:

  • r = 0$ (with multiplicity 2)

  • r = rac{-3 + ext{2024 å¹´ 8 月 17 æ—¥}}{2}

  • r = rac{-3 - ext{2024 å¹´ 8 月 17 æ—¥}}{2}

Therefore, the equation has four roots, as expected for a fourth-degree polynomial. Two of the roots are identical (r = 0), and the other two are irrational roots derived from the quadratic formula.

Conclusion

In conclusion, the most efficient way to solve the equation $4r^4 + 12r^3 + 2r^2 = 0$ is to factor by GCF and then use the quadratic formula (Option B). Factoring by GCF simplifies the equation, and the quadratic formula provides a direct method for finding the roots of the resulting quadratic expression. We found that the solutions are $r = 0$ (with multiplicity 2), $r = rac{-3 + ext{2024 年 8 月 17 日}}{2}$, and $r = rac{-3 - ext{2024 年 8 月 17 日}}{2}$. Understanding these steps not only helps in solving this particular equation but also provides a solid foundation for tackling similar polynomial equations in the future. Keep practicing, and you'll become a pro at solving these types of problems in no time! Great job, guys!