Factoring And Solving Quartic Equations

When faced with a quartic equation like $4 x^4+7 x^2-2=0$, it might seem a bit daunting at first glance. However, these types of equations often have a hidden structure that makes them much more manageable. We're going to dive into how to solve this specific equation, focusing on finding its factored form and then uncovering all four of its solutions. This process typically involves a clever substitution that transforms the quartic equation into a quadratic one, which we already know how to handle! Let's break down the steps involved in tackling $4 x^4+7 x^2-2=0$. The first crucial step is to recognize that this equation resembles a quadratic equation if we consider $x^2$ as our variable. This technique, often called a 'u-substitution' or 'quadratic form,' is a powerful tool in algebra. By letting $u = x^2$, our original equation $4 x^4+7 x^2-2=0$ transforms into $4 u^2+7 u-2=0$. Now, this looks like a standard quadratic equation, which we can solve using various methods, such as factoring or the quadratic formula. The goal here is to find the values of $u$ first. Once we have the values for $u$, we can then substitute back $x^2$ for $u$ and solve for $x$. This substitution is key because it simplifies the complexity of the quartic equation into something much more familiar and solvable. Remember, the degree of the polynomial often gives us a clue about the number of solutions we should expect. For a quartic equation (degree 4), we anticipate finding four solutions, which might include real numbers, complex numbers, or a combination of both. The factored form will also reveal these relationships. So, by transforming $4 x^4+7 x^2-2=0$ into $4 u^2+7 u-2=0$, we've already made significant progress towards unlocking its secrets and finding all its solutions.

Factoring the Quadratic Form

Now that we have transformed our quartic equation $4 x^4+7 x^2-2=0$ into the quadratic equation $4 u^2+7 u-2=0$ by substituting $u = x^2$, the next logical step is to solve for $u$. The most straightforward method for this quadratic equation is factoring. We are looking for two binomials that multiply together to give us $4 u^2+7 u-2$. Typically, we look for factors of the first term ($4u^2$) and factors of the last term (-2) and arrange them within the binomials such that their outer and inner products sum up to the middle term ($+7u$). For $4u^2$, we can have $(4u ext{ })(u ext{ })$ or $(2u ext{ })(2u ext{ })$. For -2, we can have $(+2)(-1)$ or $(-2)(+1)$. Let's try combinations to see which one yields $+7u$. If we use $(4u ext{ })(u ext{ })$ and try the factors of -2 as $(+2)(-1)$, we get $(4u+2)(u-1)$. Expanding this gives $4u^2 - 4u + 2u - 2 = 4u^2 - 2u - 2$, which is not correct. Let's try another combination. How about $(4u ext{ })(u ext{ })$ with factors $(-1)(+2)$? This gives $(4u-1)(u+2)$. Expanding this: $(4u)(u) + (4u)(2) + (-1)(u) + (-1)(2) = 4u^2 + 8u - u - 2 = 4u^2 + 7u - 2$. This is exactly our quadratic equation! So, the factored form of $4 u^2+7 u-2$ is $(4u-1)(u+2)$. Therefore, setting this factored form to zero, we have $(4u-1)(u+2)=0$. This implies either $4u-1=0$ or $u+2=0$. Solving for $u$ in each case, we get u = rac{1}{4} and $u = -2$. These are the two values for $u$. The ability to factor the quadratic form is a key step that simplifies the entire problem, allowing us to move forward to find the original variable, $x$. This process highlights the elegance of algebraic manipulation and how recognizing patterns can drastically reduce the complexity of solving equations.

Substituting Back and Solving for x

We've successfully factored the quadratic form, finding that $4u^2+7u-2=0$ factors into $(4u-1)(u+2)=0$, which gave us the solutions u = rac{1}{4} and $u = -2$. Now, the crucial part is to remember our original substitution: $u = x^2$. We need to substitute $x^2$ back in for $u$ in both of these solutions to find the values of $x$. This is where we transition from solving the quadratic auxiliary equation back to solving the original quartic equation. For the first solution, u = rac{1}{4}, we set x^2 = rac{1}{4}. To solve for $x$, we take the square root of both sides. Remember that taking the square root introduces both a positive and a negative solution. So, x = pm@rac{1}{4}, which means x = rac{1}{2} and x = -rac{1}{2}. These are two of our four solutions. For the second solution, $u = -2$, we set $x^2 = -2$. Again, we take the square root of both sides: $x = pm@{-2}$. This will give us complex solutions. The square root of -2 can be written as $pm@{-1 imes 2}$, which is $pm@{-1} imes pm@{2}$. Since $pm@{-1}$ is the imaginary unit $i$, we have $x = pm@{-2} = i pm@{2}$ and $x = -i pm@{2}$. So, our four solutions for the original equation $4 x^4+7 x^2-2=0$ are rac{1}{2}, -rac{1}{2}, $i pm@{2}$, and $-i pm@{2}$. This step is vital because it fully answers the problem by providing the values of $x$ that satisfy the original equation. It also illustrates how complex solutions arise naturally from equations involving negative numbers under square roots. The substitution technique is a powerful way to unravel complex polynomial equations into simpler ones, making them accessible for solving.

The Factored Form of the Quartic

Having found the values of $u$ and subsequently solved for $x$, we can now construct the factored form of the original quartic polynomial $4 x^4+7 x^2-2$. The factored form is directly related to the solutions we found. If a polynomial has roots $r_1, r_2, r_3, r_4$, then it can be expressed in the form $a(x-r_1)(x-r_2)(x-r_3)(x-r_4)$, where $a$ is the leading coefficient. In our case, the leading coefficient of $4 x^4+7 x^2-2$ is 4. Our four solutions for $x$ are rac{1}{2}, -rac{1}{2}, $i pm@{2}$, and $-i pm@{2}$. So, the factored form should be 4 ig(x - rac{1}{2}ig) ig(x - (-rac{1}{2})ig) ig(x - i pm@{2}ig) ig(x - (-i pm@{2})ig). This simplifies to 4 ig(x - rac{1}{2}ig) ig(x + rac{1}{2}ig) ig(x - i pm@{2}ig) ig(x + i pm@{2}ig). We can group the factors that yield real coefficients. Notice that ig(x - rac{1}{2}ig) ig(x + rac{1}{2}ig) is the difference of squares, (x^2 - (rac{1}{2})^2) = x^2 - rac{1}{4}. Also, ig(x - i pm@{2}ig) ig(x + i pm@{2}ig) is also a difference of squares, $(x^2 - (i pm@{2})^2) = x^2 - (i^2 imes 2) = x^2 - (-1 imes 2) = x^2 - (-2) = x^2+2$. So, the factored form becomes 4 ig(x^2 - rac{1}{4}ig) (x^2+2). If we distribute the 4 into the first factor, we get 4(x^2 - rac{1}{4}) = 4x^2 - 4(rac{1}{4}) = 4x^2 - 1. Thus, the factored form is ig(4x^2-1ig)(x^2+2). This matches one of the choices provided. This factored form is crucial as it directly represents the structure of the polynomial and clearly shows the origin of its roots. It's a testament to the fact that even complex roots, when paired with their conjugates, can lead to factors with real coefficients.

Summary of Solutions and Factored Form

To recap our journey in solving the quartic equation $4 x^4+7 x^2-2=0$, we employed a substitution method that proved remarkably effective. By letting $u = x^2$, we transformed the equation into a more manageable quadratic form: $4 u^2+7 u-2=0$. This quadratic equation was then factored into $(4u-1)(u+2)=0$, yielding two solutions for $u$: u = rac{1}{4} and $u = -2$. The critical next step was to substitute back $x^2$ for $u$. For u = rac{1}{4}, we solved x^2 = rac{1}{4} to find two real solutions: x = rac{1}{2} and x = -rac{1}{2}. For $u = -2$, we solved $x^2 = -2$ to find two complex solutions: $x = i pm@{2}$ and $x = -i pm@{2}$. Therefore, the four correct solutions to the equation $4 x^4+7 x^2-2=0$ are indeed rac{1}{2}, -rac{1}{2}, $i pm@{2}$, and $-i pm@{2}$. These four solutions consist of two real numbers and two complex conjugates, as is common for polynomial equations with real coefficients. Furthermore, we derived the factored form of the polynomial. Using the roots, we constructed the factored expression as 4 ig(x - rac{1}{2}ig) ig(x + rac{1}{2}ig) ig(x - i pm@{2}ig) ig(x + i pm@{2}ig). By grouping conjugate pairs and simplifying, we arrived at the factored form ig(4x^2-1ig)(x^2+2). This factored form is elegant because it shows how the original quartic polynomial can be broken down into simpler quadratic factors, with one factor yielding real roots and the other yielding complex roots. The entire process demonstrates the power of substitution and the fundamental theorem of algebra, which guarantees that a polynomial of degree $n$ has exactly $n$ roots (counting multiplicity) in the complex number system. Understanding these techniques is fundamental for anyone delving deeper into algebra and calculus.

For further exploration into the fascinating world of algebraic equations and polynomial factoring, you might find the resources at MathWorld to be incredibly insightful. Their comprehensive articles cover a vast array of mathematical topics with depth and clarity.