Solving Equations Reducible to Quadratics: Techniques and Examples

Equations that can be transformed into quadratic form are a common type of problem encountered in algebra. By introducing appropriate substitution, these equations can be simplified and solved systematically. This article will explore the techniques and solutions for a pair of equations that are reducible to quadratic equations, along with some observations about their complex nature. We will also discuss the use of metavariables to streamline the solving process.

Introduction to Reducible Equations

Equations such as those in the form ( ax^m bx^n c ) can often be simplified by introducing a new variable that converts them into a quadratic equation. This technique is particularly useful when dealing with higher-degree polynomials. The substitution method helps in reducing the complexity and makes the problem more manageable.

Example 1: x^1/3 - x^2/3 5

Let the equation be ( x^{1/3} - x^{2/3} 5 ). To simplify this equation, we can use the substitution:

Let ( u x^{1/3} ). Then, the equation becomes ( u - u^2 5 ). Rearrange to form a standard quadratic equation: ( u^2 - u 5 0 ). Solve using the quadratic formula. The discriminant of the quadratic equation ( u^2 - u 5 0 ) is [ Delta (-1)^2 - 4(1)(5) 1 - 20 -19 ]. Since the discriminant is negative, the equation has no real solutions. However, we can find the complex solutions as follows: Using the quadratic formula: [ u frac{-(-1) pm sqrt{(-19)}}{2(1)} frac{1 pm isqrt{19}}{2} ]. Therefore, the solutions for ( u ) are ( u frac{1 isqrt{19}}{2} ) and ( u frac{1 - isqrt{19}}{2} ). Recall that ( u x^{1/3} ), so: [ x u^3 ]. Substituting back, we find the complex solutions for ( x ).

Example 2: 21t^-218t^-4 9

Let's solve the equation ( 21t^{-2} - 18t^{-4} 9 ):

Introduce the substitution ( u t^{-2} ), so the equation becomes ( 21u - 18u^2 9 ). Divide through by 3 to simplify: ( 7u - 6u^2 3 ). Multiply through by (-1) for convenience: (-6u^2 7u - 3 0 ). Rearrange to form a standard quadratic equation: ( 6u^2 - 7u 3 0 ). Solve the quadratic equation using the quadratic formula: [ u frac{-(-7) pm sqrt{(-7)^2 - 4(6)(3)}}{2(6)} frac{7 pm sqrt{49 - 72}}{12} frac{7 pm sqrt{-23}}{12} frac{7 pm isqrt{23}}{12} ]. Thus, the solutions for ( u ) are ( u frac{7 isqrt{23}}{12} ) and ( u frac{7 - isqrt{23}}{12} ). Recall that ( u t^{-2} ), so: [ t^2 frac{1}{u} ]. Substituting back, we find the complex solutions for ( t ).

General Techniques and Metavariables

Using metavariables can be a powerful tool in solving complex equations. By introducing a new variable, the equation can often be reduced to a quadratic form, making it easier to solve. For example, in the given equations:

For the first equation, we can set ( u x^{1/3} ) to simplify the equation to a quadratic in ( u ). For the second equation, we can use ( u t^{-2} ) to transform it into a quadratic form.

The use of metavariables can simplify the solving process and make it more intuitive. However, it is essential to substitute back into the original variables to find the final solutions accurately.

It's important to note that while these equations can be solved, they may have complex solutions, especially if the discriminant is negative. In such cases, the solutions involve imaginary numbers, as seen in the examples above.

Conclusion

Understanding the methods of solving equations reducible to quadratics is crucial for solving a wide range of algebraic problems. The use of substitutions and metavariables can simplify the solving process and provide insight into the nature of the solutions. Whether you are dealing with real or complex solutions, mastering these techniques will greatly improve your problem-solving skills in algebra.