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Magazine Chapter 11: Problem 25
Solve each equation by making an appropriate substitution. If at any point in the solution process both sides of an equation are raised to an even power, a check is required. $$x^{\frac{2}{3}}-x^{\frac{1}{3}}-6=0$$
Short Answer
Expert verified
The solution to the equation \(x^{\frac{2}{3}}-x^{\frac{1}{3}}-6=0\) is \(x = 27\)
Step by step solution
01
Substitution
Substitute \(x^{\frac{1}{3}}\) with \(y\). It transforms our equation into the quadratic form: \(y^2-y-6=0\)
02
Solve the Quadratic Equation
Now, we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. The equation factors to \((y-3)(y+2) = 0\) which indicates that the solutions in terms of \(y\) are \(y = 3\) or \(y = -2\)
03
Substitute Back
Replace \(y\) with \(x^{\frac{1}{3}}\), giving us the equations \(x^{\frac{1}{3}} = 3\) and \(x^{\frac{1}{3}} = -2\)
04
Solve for x
To get \(x\), we can cube both sides. Solving these two equations gives \(x = 27\) for the first equation and \(x = -8\) for the second equation.
05
Recheck
As a check, substitute these values back into the original equation to see if these satisfy the initial equation. \(x=27\) passes the check while \(x=-8\) does not satisfy the condition as it leads to a negative number under a cube root.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful algebraic technique that simplifies complex equations by introducing a new variable. This is particularly helpful for equations involving exponents or fractional powers, which can be intimidating at first glance.
- In the original exercise, we encountered an equation with fractional exponents. By substituting a part of the equation with a new variable, specifically replacing \(x^{\frac{1}{3}}\) with \(y\), we transformed a seemingly complicated expression into a more familiar quadratic form.
- This makes it much easier to manage and solve, as quadratics are a standard component of algebra curricula and come with established techniques for solving them.
Quadratic Equation
Quadratic equations are polynomial equations of the second degree, typically expressed in the form \(ax^2 + bx + c = 0\). These equations are crucial because they appear frequently in various mathematical contexts.
Solving a quadratic equation often involves finding the values of the variable that make the equation true. There are several methods to solve quadratic equations:
Solving a quadratic equation often involves finding the values of the variable that make the equation true. There are several methods to solve quadratic equations:
- Factoring: This involves rewriting the quadratic in a product form, as we did in the original solution. Our example \(y^2-y-6=0\) factors to \((y-3)(y+2) = 0\).
- Completing the Square: A method that involves creating a perfect square trinomial from the original quadratic equation.
- Quadratic Formula: A universal tool \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) used when factoring is not feasible or is too complex.
Cubic Roots
Cubic roots involve determining a number which, when raised to the power of three, gives the original number. In our exercise, once we have solved for \(y\) and substituted it back with \(x^{\frac{1}{3}}\), we need to resolve these back to \(x\).
- The relationship stems from the power property of exponents, namely that \((x^{m})^n = x^{mn}\). Cube root operations apply directly when you raise the expression to the power of three to return to the original number.
- When we find that \(x^{\frac{1}{3}} = 3\), cubing both sides gives \(x = 3^3 = 27\).
- Handling negative numbers under cubic roots requires special attention, as negative base values yield negative results when cubed. In this instance, not all solutions, like \(x^{\frac{1}{3}} = -2\), work under even-powered transformations or real number constraints.
Factoring Quadratic Equations
Factoring quadratic equations simplifies the process of finding the roots of a polynomial. It involves breaking down the quadratic into simpler expressions that, when multiplied together, give the original equation.
- In our problem, the quadratic \(y^2-y-6=0\) is factored into \((y-3)(y+2)=0\). This step indicates that \(y = 3\) or \(y = -2\) are the solutions because, for any product to be zero, at least one of its multiplicands must be zero.
- Successful factoring hinges on recognizing patterns, such as the difference of squares or perfect square trinomials, and utilizing common techniques to rewrite expressions.
- It also provides a deeper insight into the properties of polynomial equations and makes solving them less cumbersome through simpler arithmetic operations.
