Chapter 0: Problem 59
Solve the equations by introducing a substitution that transforms these equations to quadratic form. $$z^{2 / 5}-2 z^{1 / 5}+1=0$$
Short Answer
Expert verified
The solution is \( z = 1 \).
Step by step solution
01
Identify Suitable Substitution
We have the equation \( z^{2/5} - 2z^{1/5} + 1 = 0 \). To convert it to a standard quadratic form, notice the terms \( z^{2/5} \) and \( z^{1/5} \). We can use a substitution where \( t = z^{1/5} \). Then, \( z^{2/5} \) becomes \( t^2 \).
02
Substitute and Simplify the Equation
Apply the substitution to the equation, replacing \( z^{1/5} \) with \( t \) and \( z^{2/5} \) with \( t^2 \). The equation becomes: \( t^2 - 2t + 1 = 0 \). This is now in a quadratic form.
03
Solve the Quadratic Equation
The quadratic equation \( t^2 - 2t + 1 = 0 \) can be solved by factoring. Notice it is a perfect square: \( (t - 1)^2 = 0 \). Thus, \( t = 1 \) is the solution.
04
Reverse the Substitution
Recall the substitution \( t = z^{1/5} \). We found \( t = 1 \), so we solve \( z^{1/5} = 1 \). This only means \( z = 1^5 = 1 \). Hence, \( z = 1 \).
05
Confirm the Solution
Substitute \( z = 1 \) back into the original equation to ensure it satisfies it. \( (1)^{2/5} - 2(1)^{1/5} + 1 = 1 - 2 + 1 = 0 \), which is correct.
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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 useful technique for transforming complex equations into a more manageable form. In the context of quadratic equations, this technique can often make an equation easier to solve by introducing a new variable. In our example, we started with the equation \( z^{2/5} - 2z^{1/5} + 1 = 0 \). This equation contains terms in the form of fractional exponents of \( z \), which complicates solving directly. To simplify, we identify a common term that can be represented by a single variable.
By letting \( t = z^{1/5} \), the equation simplifies because every power of \( z \) is expressible in terms of \( t \). Specifically, \( z^{2/5} \) becomes \( t^2 \). This substitution converts the original equation into a standard quadratic equation: \( t^2 - 2t + 1 = 0 \).
This method simplifies your work and makes the equation much more approachable. Once solved, as we find the value for \( t \), we can reverse the substitution to get back to a solution in terms of \( z \). This back-and-forth approach shows how substitution can simplify and solve complex problems effectively.
By letting \( t = z^{1/5} \), the equation simplifies because every power of \( z \) is expressible in terms of \( t \). Specifically, \( z^{2/5} \) becomes \( t^2 \). This substitution converts the original equation into a standard quadratic equation: \( t^2 - 2t + 1 = 0 \).
This method simplifies your work and makes the equation much more approachable. Once solved, as we find the value for \( t \), we can reverse the substitution to get back to a solution in terms of \( z \). This back-and-forth approach shows how substitution can simplify and solve complex problems effectively.
Perfect Square Trinomial
A perfect square trinomial is a quadratic expression that can be expressed as the square of a binomial. Recognizing this can greatly simplify the process of solving quadratic equations. In our example, after substitution, we have the quadratic equation \( t^2 - 2t + 1 = 0 \).
To determine if it is a perfect square trinomial, we check if it fits the form \((a-b)^2 = a^2 - 2ab + b^2 \). Here, we realize that \( (t-1)^2 \) gives us exactly \( t^2 - 2t + 1 \). Thus, it is indeed a perfect square trinomial.
This revelation simplifies solving because instead of having to use the quadratic formula, factoring, or completing the square, we can directly equate the squared term to zero: \((t-1)^2 = 0 \). Solving this allows us to find that \( t = 1 \). Recognizing and using perfect square trinomials can streamline solving quadratic equations.
To determine if it is a perfect square trinomial, we check if it fits the form \((a-b)^2 = a^2 - 2ab + b^2 \). Here, we realize that \( (t-1)^2 \) gives us exactly \( t^2 - 2t + 1 \). Thus, it is indeed a perfect square trinomial.
This revelation simplifies solving because instead of having to use the quadratic formula, factoring, or completing the square, we can directly equate the squared term to zero: \((t-1)^2 = 0 \). Solving this allows us to find that \( t = 1 \). Recognizing and using perfect square trinomials can streamline solving quadratic equations.
Factoring Quadratics
Factoring is a fundamental method utilized to solve quadratic equations. It involves expressing the quadratic equation in a form where it is the product of two binomials. This technique is invaluable since it allows for quick and straightforward solutions.
For the quadratic \( t^2 - 2t + 1 = 0 \), factoring reveals that it is a perfect square trinomial, as previously discussed. This means it factors neatly into \((t-1)(t-1) = 0 \) or equivalently \((t-1)^2 = 0 \).
The factored form provides solutions with ease. From \((t-1)^2 = 0 \), we instantaneously ascertain that \( t = 1 \). This method is efficient, especially when a quadratic can be recognized as a perfect square or factors nicely.
For the quadratic \( t^2 - 2t + 1 = 0 \), factoring reveals that it is a perfect square trinomial, as previously discussed. This means it factors neatly into \((t-1)(t-1) = 0 \) or equivalently \((t-1)^2 = 0 \).
The factored form provides solutions with ease. From \((t-1)^2 = 0 \), we instantaneously ascertain that \( t = 1 \). This method is efficient, especially when a quadratic can be recognized as a perfect square or factors nicely.
- Factorizing allows breaking down seemingly complex expressions into more manageable parts.
- Always consider factoring first for quadratics as it can reveal quick solutions.