Best Substitution For Rewriting 4x^4-21x^2+20=0 As A Quadratic Equation

Solving equations can sometimes feel like navigating a maze, but the right techniques can light the way. When faced with a quartic equation – that is, an equation where the highest power of the variable is 4 – it might seem daunting at first. However, there's often a clever trick we can employ: substitution. This involves replacing a more complex expression with a simpler variable, transforming the original equation into a more manageable form. In this article, we'll dive deep into the specific quartic equation $4x^4 - 21x^2 + 20 = 0$, explore the best substitution method to simplify it, and turn it into a quadratic equation, which we all know and love.

Understanding the Quartic Equation

Before we jump into the solution, let's break down the equation we're dealing with: $4x^4 - 21x^2 + 20 = 0$.

• Notice the pattern in the exponents: we have a term with $x^4$, a term with $x^2$, and a constant term. This pattern is key! It suggests that we can manipulate this equation into a quadratic form.
• Think of a quadratic equation: It generally looks like $ax^2 + bx + c = 0$. Our goal is to make our quartic equation resemble this form. This is where the magic of substitution comes in handy. By choosing the right substitution, we can effectively "hide" the higher powers of x and reveal a hidden quadratic structure. The coefficients 4, -21, and 20 play a crucial role in determining the roots of the equation, but our immediate focus is on the structure, not the solution itself.
• This is a crucial step to appreciate the elegance of mathematical problem-solving because recognizing patterns is fundamental to solving mathematical problems. This skill helps in simplifying complex equations and can be broadly applied in various mathematical contexts. A keen eye for patterns not only makes problem-solving more efficient but also more intuitive, paving the way for more advanced mathematical concepts and techniques. So, let's keep our eyes peeled for those patterns, as they are the keys to unlocking mathematical mysteries.

The Power of Substitution: Transforming Quartics into Quadratics

The core idea behind using substitution is to simplify complex equations by replacing a part of the equation with a single variable. This makes the equation easier to handle and solve. In our case, the presence of both $x^4$ and $x^2$ terms strongly suggests a particular substitution. We need to find a substitution that will transform $x^4$ into a squared term and $x^2$ into a linear term, effectively turning our quartic into a quadratic. Think of it like this: we're performing a sort of algebraic alchemy, transmuting a complex equation into a simpler one.

• The key to successful substitution lies in identifying the repeating element within the equation. In $4x^4 - 21x^2 + 20 = 0$, the repeating element is clearly related to $x^2$. Why? Because $x^4$ can be expressed as $(x^2)^2$. This is a crucial observation.
• Let's consider the options given: $u = x^2$, $u = 2x^2$, $u = x^4$, and $u = 4x^4$. We need to determine which of these substitutions will best serve our purpose of transforming the equation into a quadratic form. To do this, we'll test each option conceptually to see how it would affect the equation.
• If we let $u = x^2$, then $u^2 = (x^2)^2 = x^4$. This substitution neatly transforms our equation into a quadratic form in terms of u. The other options, while mathematically valid, don't lead as directly to a quadratic equation. For instance, if we use $u=x^4$, we would need to take square roots to get $x^2$ and introduce more complexity. This is the beauty of substitution: it allows us to manipulate the equation into a form that is not only more recognizable but also more solvable. By carefully choosing our substitution, we can significantly simplify the problem-solving process. The goal of substitution in this context is to simplify the equation into a more manageable form, specifically a quadratic, which we can then solve using standard methods like factoring, completing the square, or the quadratic formula. This approach not only makes the problem solvable but also highlights the power of algebraic manipulation in simplifying complex mathematical expressions.

Analyzing the Substitution Options

Let's carefully examine each of the provided substitution options to determine which one will successfully transform our quartic equation into a quadratic equation:

• Option A: $u = x^2$ This is the most promising option. If we substitute $u$ for $x^2$, then $u^2$ will be equal to $(x^2)^2$, which simplifies to $x^4$. This substitution directly addresses the relationship between the $x^4$ and $x^2$ terms in our original equation. Substituting $u$ for $x^2$ in the equation $4x^4 - 21x^2 + 20 = 0$ gives us $4u^2 - 21u + 20 = 0$. This is a quadratic equation in terms of $u$, exactly what we were aiming for! This substitution neatly transforms the quartic equation into a quadratic form, making it much easier to solve. The choice of $u = x^2$ leverages the inherent structure of the quartic equation, where $x^4$ is simply the square of $x^2$. By making this substitution, we essentially "unmask" the quadratic form hidden within the quartic, allowing us to apply familiar techniques for solving quadratic equations. This is a powerful illustration of how strategic substitutions can simplify complex mathematical problems.
• Option B: $u = 2x^2$ While this substitution might seem similar to option A, it introduces an extra coefficient that could complicate things slightly. If we substitute $u = 2x^2$, then $x^2 = u/2$. Therefore, $x^4 = (x^2)^2 = (u/2)^2 = u^2/4$. Substituting these into our original equation, we get $4(u^2/4) - 21(u/2) + 20 = 0$, which simplifies to $u^2 - (21/2)u + 20 = 0$. While this is still a quadratic equation, the fractional coefficient might make it a bit less straightforward to solve compared to option A. The introduction of the coefficient '2' in the substitution leads to a fractional coefficient in the resulting quadratic equation, which can make manual solving more cumbersome. Although it doesn't fundamentally change the solvability of the equation, it adds an extra layer of arithmetic complexity. This highlights an important aspect of mathematical problem-solving: the goal is not just to find a solution, but to find the most efficient solution. Option A, with its simpler substitution, directly leads to a quadratic equation with integer coefficients, making it the preferred approach.
• Option C: $u = x^4$ If we try this substitution, we get $4u - 21 ext{sqrt}(u) + 20 = 0$. This is not a quadratic equation. It includes a square root term, which makes it more complex to solve. This substitution, while seemingly intuitive, actually moves us further away from our goal of a quadratic form. The introduction of the square root term fundamentally changes the nature of the equation, making it a non-quadratic equation. This outcome underscores the importance of carefully considering the implications of a substitution before applying it. The best substitutions are those that simplify the equation while maintaining its essential structure, allowing us to apply known solution methods. In this case, the substitution $u = x^4$ leads to an equation that is more difficult to solve than the original, highlighting the need for strategic thinking in mathematical problem-solving.
• Option D: $u = 4x^4$ Similar to option C, this substitution also doesn't lead to a clean quadratic equation. We'd have to deal with square roots again. Let's see what happens: If $u = 4x^4$, then $x^4 = u/4$. To find $x^2$, we'd need to take the square root, leading to the same issues as with option C. This substitution, like option C, introduces complexities rather than simplifying the equation. By substituting $u = 4x^4$, we encounter similar difficulties as with $u = x^4$, primarily the need to deal with square roots to express the original equation in terms of $u$. This results in a non-quadratic form that is more challenging to solve. The lesson here is that not all substitutions are created equal. The effectiveness of a substitution depends on how well it aligns with the inherent structure of the equation, and in this case, both options C and D fail to capture the underlying quadratic nature of the quartic equation.

The Winning Substitution: $u = x^2$

As we've analyzed, the substitution $u = x^2$ (Option A) is the clear winner. It directly transforms the quartic equation $4x^4 - 21x^2 + 20 = 0$ into the quadratic equation $4u^2 - 21u + 20 = 0$. This new quadratic equation is much easier to solve using standard techniques like factoring, the quadratic formula, or completing the square. By making this substitution, we effectively reduce the complexity of the problem, allowing us to apply familiar methods to find the solutions. This highlights the power of algebraic manipulation in simplifying complex mathematical expressions and making them more accessible. This substitution effectively bridges the gap between quartic and quadratic equations, demonstrating how a strategic choice of variables can dramatically simplify the problem-solving process. The resulting quadratic equation is a testament to the elegance and efficiency of this substitution.

Solving the Quadratic Equation

Now that we've successfully transformed the quartic equation into a quadratic equation, let's solve it! We have $4u^2 - 21u + 20 = 0$. We can solve this quadratic equation by factoring. Factoring involves finding two binomials that, when multiplied together, give us the quadratic expression. We look for two numbers that multiply to give $(4)(20)=80$ and add up to -21. Those numbers are -16 and -5.

• Rewrite the middle term: $4u^2 - 16u - 5u + 20 = 0$
• Factor by grouping: $4u(u - 4) - 5(u - 4) = 0$
• Factor out the common binomial: $(4u - 5)(u - 4) = 0$
• Set each factor equal to zero: $4u - 5 = 0$ or $u - 4 = 0$
• Solve for u: $u = 5/4$ or $u = 4$

Back-Substitution: Finding the Values of x

Remember, we're not actually interested in the values of $u$; we want to find the values of $x$! So, we need to substitute back using our original substitution, $u = x^2$. This is a crucial step because we've only solved for our temporary variable, $u$. To find the solutions for the original variable, $x$, we must reverse the substitution process.

• Case 1: $u = 5/4$ Substitute back into $u = x^2$: $x^2 = 5/4$. Take the square root of both sides: $x = ± ext{sqrt}(5/4) = ± ext{sqrt}(5)/2$. Don't forget the ± sign! Square roots have both positive and negative solutions. This is a common mistake, but remembering to consider both roots is crucial for finding all possible solutions to the original equation. The positive and negative roots arise from the fact that squaring either a positive or a negative number yields a positive result. Therefore, when reversing the square operation, we must account for both possibilities.
• Case 2: $u = 4$ Substitute back into $u = x^2$: $x^2 = 4$. Take the square root of both sides: $x = ±2$. Again, remember the ± sign! This highlights the fundamental property of square roots and the importance of considering both positive and negative solutions. When we solve $x^2 = 4$, we're essentially asking, "What numbers, when squared, equal 4?" Both 2 and -2 satisfy this condition, hence the two solutions. The recognition of both solutions is vital for a complete and accurate understanding of the equation's behavior.

The Solutions to the Quartic Equation

Therefore, the solutions to the quartic equation $4x^4 - 21x^2 + 20 = 0$ are $x = ext{sqrt}(5)/2$, $x = - ext{sqrt}(5)/2$, $x = 2$, and $x = -2$. We've successfully navigated the quartic maze by using the clever technique of substitution! This journey demonstrates the power of algebraic manipulation and the beauty of recognizing patterns in mathematical equations. By transforming a complex quartic equation into a more familiar quadratic form, we were able to apply standard solution methods and arrive at the complete set of solutions. This approach not only simplifies the problem-solving process but also highlights the interconnectedness of different mathematical concepts and techniques. The ability to see these connections and leverage them is a hallmark of mathematical fluency and a key to success in solving complex problems.

Conclusion

The substitution $u = x^2$ is the key to rewriting the quartic equation $4x^4 - 21x^2 + 20 = 0$ as a quadratic equation. This technique allows us to transform a seemingly complex problem into a more manageable form, highlighting the power of algebraic manipulation. Remember, guys, when faced with a challenging equation, look for patterns and consider the magic of substitution! By strategically choosing the right substitution, we can unlock the hidden structure of the equation and make it much easier to solve. This approach not only simplifies the problem-solving process but also deepens our understanding of the underlying mathematical principles. So, keep practicing, keep exploring, and keep embracing the power of substitution in your mathematical journey!