Solving Quadratic Equations in Quadratic Form: A Step-by-Step Guide
Solving equations in quadratic form can often be simplified into a more manageable problem through a clever substitution. Specifically, when a polynomial equation has a term that is a perfect square, we can substitute a variable to transform the equation into a standard quadratic form. In this guide, we'll walk you through the process of solving the equation 4x4-5x2 1 0 using substitution and the quadratic formula.
Understanding the Base Equation
The equation we are dealing with is 4x4-5x2 1 0. This equation is in the quadratic form with the highest power of x being 4, but it can be made simpler by recognizing that x2 is a common term. This is a tell-tale sign that substitution might be the way to go.
The Substitution Method
Let's set a substitution: y x2. The key idea here is that we are changing the form of the equation while preserving its essential properties. Substituting y into our equation gives us:
4y2-5y 1 0
This is now in the standard quadratic form, where all the terms are in relation to y.
Solving the New Quadratic Equation
Now that we have a standard quadratic equation, we can solve it using methods like the quadratic formula, factoring, or completing the square. In this case, we can solve it through factoring:
4y2-5y 1 4y2-4y-y 1 4y(y-1)-1(y-1) (4y-1)(y-1) 0
Setting each factor equal to zero gives us:
4y - 1 0 or y - 1 0
Solving these, we find:
y 1/4 or y 1
Back to the Original Variable
Recall that y x2. Therefore, substituting our solution back into this gives us:
x2 1/4 or x2 1
Now, we solve for x by taking the square root of both sides:
x ±1/2 or x ±1
Graphical Interpretation and Real-World Applications
Graphically, the roots of the equation y x2-5x 1 represent the x-intercepts of the parabola defined by this function. These points are significant in many real-world applications, such as projectile motion, where the equation describes the height (h) of the object as a function of time (t), and the roots correspond to the times when the object is at ground level.
Conclusion
By using the substitution method, we were able to transform a complex polynomial equation into a more straightforward quadratic equation, making it easier to solve. This approach is not only useful for solving such equations but also serves as a foundational technique in more advanced mathematical problems. Understanding this process can be beneficial for high school and college students, as well as professionals in fields that require a strong grasp of algebraic manipulations.