Rewrite 4x^4 - 21x^2 + 20 = 0 As A Quadratic Equation

Have you ever looked at a complex equation and wished there was a simpler way to tackle it? Well, in the realm of mathematics, especially when dealing with polynomial equations, there often is! One powerful technique that can make solving higher-degree polynomials much more manageable is the use of substitution. This method allows us to transform an equation that might seem intimidating at first glance into a form we're already familiar with, like a quadratic equation. Let's dive into how we can rewrite a specific quartic equation, 4x⁴ - 21x² + 20 = 0, into a quadratic form and explore the best substitution to achieve this. Understanding this process is key to unlocking solutions for a wider range of algebraic problems.

The Power of Substitution in Algebra

The core idea behind substitution in algebra is to replace a complex expression with a simpler variable, often denoted by a letter like 'u' or 'y'. This is particularly useful when an equation contains terms that are powers of another term. In our target equation, 4x⁴ - 21x² + 20 = 0, we can observe a clear relationship between the terms involving 'x'. Notice that x⁴ is actually the square of x². Specifically, x⁴ = (x²)². This relationship is the crucial hint that substitution will be effective here. When we spot this pattern, we can introduce a new variable to represent the 'base' term, which in this case is x². By doing this, we essentially simplify the exponents, turning a degree-four equation into a degree-two equation – a quadratic! This transformation is incredibly valuable because we have well-established methods, such as factoring, the quadratic formula, or completing the square, to solve quadratic equations. Without substitution, solving a quartic equation directly can be significantly more challenging, often requiring more advanced techniques.

Identifying the Correct Substitution

To effectively rewrite 4x⁴ - 21x² + 20 = 0 as a quadratic equation, our substitution needs to address the relationship between x⁴ and x². Let's consider the options provided:

• A. u = 2x²: If we choose this substitution, then x² = u/2. Substituting this into the original equation would give us 4(u/2)² - 21(u/2) + 20 = 0. This simplifies to 4(u²/4) - 21u/2 + 20 = 0, which further reduces to u² - (21/2)u + 20 = 0. While this does result in a quadratic equation in terms of 'u', the coefficients are fractions, which might be less convenient than other possibilities. More importantly, it doesn't directly represent the structure of the original equation in the most straightforward way.

• B. u = x²: This is where the magic happens! If we set u = x², then by the laws of exponents, x⁴ = (x²)² = u². Now, let's substitute these into the original equation: 4(u²) - 21(u) + 20 = 0. Look at that! We have transformed the original quartic equation into a standard quadratic equation in terms of 'u': 4u² - 21u + 20 = 0. This substitution perfectly captures the relationship between the x⁴ and x² terms, resulting in a clean and familiar quadratic form. This is generally the most elegant and direct substitution for this type of problem.

• C. u = x⁴: If we let u = x⁴, then we would need to express x² in terms of u. Since x⁴ = (x²)², it implies that x² = √(x⁴) = √u (assuming x² is non-negative, which is often the case when dealing with real solutions, but it introduces complexity). Substituting this would lead to 4u - 21√u + 20 = 0. This is not a quadratic equation; it's an equation involving a square root, which is a different kind of problem altogether.

• D. u = 4x⁴: Similar to option C, if u = 4x⁴, then x⁴ = u/4. To substitute x², we'd again need to involve square roots, leading to x² = √(x⁴) = √(u/4). The equation would become 4(u/4) - 21√(u/4) + 20 = 0, simplifying to u - (21/2)√u + 20 = 0. Again, this is not a quadratic equation and is more complicated than necessary.

Based on this analysis, the substitution u = x² (Option B) is the most appropriate choice because it directly transforms the quartic equation 4x⁴ - 21x² + 20 = 0 into a standard quadratic equation 4u² - 21u + 20 = 0, where u represents x². This simplification makes the equation much easier to solve using standard algebraic techniques.

Solving the Transformed Quadratic Equation

Once we have successfully rewritten the equation 4x⁴ - 21x² + 20 = 0 as the quadratic equation 4u² - 21u + 20 = 0 using the substitution u = x², we can proceed to solve for 'u'. There are several methods to solve quadratic equations, but factoring and the quadratic formula are the most common. Let's try factoring first, as it often provides a quicker solution if the quadratic is factorable.

We are looking for two binomials that multiply to give 4u² - 21u + 20. We need two numbers that multiply to (4 * 20) = 80 and add up to -21. Let's list pairs of factors of 80 and check their sums:

• 1 and 80 (sum 81)
• 2 and 40 (sum 42)
• 4 and 20 (sum 24)
• 5 and 16 (sum 21)
• 8 and 10 (sum 18)

Since we need the sum to be -21, and the product is positive, both numbers must be negative. Thus, we are looking for two negative numbers that multiply to 80 and add to -21. The pair -5 and -16 fits this description perfectly, as (-5) * (-16) = 80 and (-5) + (-16) = -21.

Now, we can use these numbers to rewrite the middle term (-21u) and factor by grouping:

4u² - 21u + 20 = 0 4u² - 16u - 5u + 20 = 0

Group the terms:

(4u² - 16u) + (-5u + 20) = 0

Factor out the greatest common factor from each group:

4u(u - 4) - 5(u - 4) = 0

Now, factor out the common binomial (u - 4):

(4u - 5)(u - 4) = 0

For this product to be zero, one or both of the factors must be zero:

• 4u - 5 = 0 => 4u = 5 => u = 5/4
• u - 4 = 0 => u = 4

So, the solutions for 'u' are u = 5/4 and u = 4.

Back-Substitution: Finding the Values of x

Remember, our goal was to solve for 'x', not 'u'. We used the substitution u = x². Now we need to substitute back the values of 'u' we found to determine the corresponding values of 'x'.

Case 1: u = 5/4

If u = 5/4, then x² = 5/4. To solve for x, we take the square root of both sides: x = ±√(5/4) x = ±(√5 / √4) x = ±√5 / 2

So, two solutions for x are x = √5 / 2 and x = -√5 / 2.

Case 2: u = 4

If u = 4, then x² = 4. Taking the square root of both sides: x = ±√4 x = ±2

So, two more solutions for x are x = 2 and x = -2.

Therefore, the original quartic equation 4x⁴ - 21x² + 20 = 0 has four solutions: x = 2, x = -2, x = √5 / 2, and x = -√5 / 2. The substitution u = x² was instrumental in simplifying the problem and allowing us to find these solutions efficiently.

Why Other Substitutions Fail

It's worth reiterating why the other substitution options, while superficially related to the equation, don't lead to a direct quadratic form. When we aim to rewrite an equation like 4x⁴ - 21x² + 20 = 0 as a quadratic, we are specifically looking for a substitution u = f(x) such that the equation transforms into a*u² + b*u + c = 0. This structure implies that the original equation must be expressible in terms of f(x) and [f(x)]².

In our case, the terms are x⁴ and x². The key observation is that x⁴ = (x²)². This means if we let our substitution u be x², then u² will naturally be (x²)² = x⁴. This creates the perfect u² and u relationship needed for a quadratic equation.

If we tried u = 2x² (Option A), we get x² = u/2 and x⁴ = (u/2)² = u²/4. The equation becomes 4(u²/4) - 21(u/2) + 20 = 0, which is u² - (21/2)u + 20 = 0. While this is a quadratic equation, it's not the simplest or most direct transformation. It introduces fractional coefficients that we might prefer to avoid if possible. The standard approach is to make the substitution for the lowest power term that, when squared, gives the highest power term. In x⁴ and x², x² is that term.

Options C (u = x⁴) and D (u = 4x⁴) fail because they don't allow us to express the x² term in a simple polynomial form of u. As shown earlier, x² would involve a square root of u (or a related term), leading to radical equations, not quadratic ones. For instance, if u = x⁴, then x² = √u (for real, non-negative x²). Substituting this into 4x⁴ - 21x² + 20 = 0 yields 4u - 21√u + 20 = 0. This equation is fundamentally different from a quadratic equation and requires different solving techniques.

Therefore, the choice of substitution is critical. It must align perfectly with the powers of the variable present in the equation to achieve the desired transformation into a quadratic form. For 4x⁴ - 21x² + 20 = 0, the perfect fit is u = x².

Conclusion: The Elegance of Algebraic Transformation

In conclusion, when faced with polynomial equations where higher powers are simply squares of lower powers, the technique of substitution offers an elegant pathway to simplification. For the specific equation 4x⁴ - 21x² + 20 = 0, the most effective substitution to rewrite it as a quadratic equation is u = x². This choice directly transforms the quartic into the quadratic form 4u² - 21u + 20 = 0, which can then be solved using standard methods like factoring or the quadratic formula. The solutions obtained for 'u' are then back-substituted to find the final solutions for 'x'. Mastering this technique not only helps in solving specific problems but also builds a deeper understanding of the structure and relationships within algebraic expressions. It's a testament to the power of variable manipulation in making complex mathematical challenges more accessible.

For further exploration into solving polynomial equations and the techniques of substitution, you can visit reliable resources like ** Khan Academy.**