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Magazine Chapter 1: Problem 61
Find the real solutions of each equation. $$ x^{1 / 2}-3 x^{1 / 4}+2=0 $$
Short Answer
Expert verified
The real solutions are \( x = 1 \) and \( x = 16 \).
Step by step solution
01
Substitute to simplify
Let鈥檚 use the substitution method to simplify the equation. Let鈥檚 set \( y = x^{1/4} \). Therefore, \( y^2 = x^{1/2} \). Substituting these into the equation gives: \[ y^2 - 3y + 2 = 0 \]
02
Factor the quadratic equation
Next, factor the quadratic equation \( y^2 - 3y + 2 = 0 \). We need two numbers that multiply to 2 and add up to -3. These numbers are -1 and -2. Thus, the factored equation is: \[ (y - 1)(y - 2) = 0 \]
03
Solve for y
Set each factor to zero and solve for \( y \): \[ y - 1 = 0 \, \Rightarrow \, y = 1 \] \[ y - 2 = 0 \, \Rightarrow \, y = 2 \]
04
Substitute back to solve for x
Now we need to substitute back \( y = x^{1/4} \) to find the values of \( x \): When \( y = 1 \), \[ x^{1/4} = 1 \] Raising both sides to the 4th power, \[ x = 1 \] When \( y = 2 \), \[ x^{1/4} = 2 \] Raising both sides to the 4th power, \[ x = 16 \]
05
State the solutions
The solutions to the original equation are: \( x = 1 \) and \( x = 16 \)
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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 way to simplify equations, especially when dealing with complex expressions. By introducing a new variable, we can transform a complex equation into a simpler one.
In the given problem, we introduced a variable, let's call it \(y\), to simplify the expression. We set \(y = x^{1/4}\), which means \(y^2 = x^{1/2}\). This substitution made the original equation easier to manipulate:
In the given problem, we introduced a variable, let's call it \(y\), to simplify the expression. We set \(y = x^{1/4}\), which means \(y^2 = x^{1/2}\). This substitution made the original equation easier to manipulate:
- Original Equation: \(x^{1/2} - 3x^{1/4} + 2 = 0\)
- After Substitution: \(y^2 - 3y + 2 = 0\).
Quadratic Equation
A quadratic equation generally takes the form \(ax^2 + bx + c = 0\). Here, our simplified equation \(y^2 - 3y + 2 = 0\) is a quadratic equation.
Quadratic equations can be solved using a variety of methods, such as factoring, completing the square, or using the quadratic formula. Understanding the structure of a quadratic equation is essential because it often appears in high school math and beyond.
In our context, the coefficients are:
Quadratic equations can be solved using a variety of methods, such as factoring, completing the square, or using the quadratic formula. Understanding the structure of a quadratic equation is essential because it often appears in high school math and beyond.
In our context, the coefficients are:
- a = 1 (coefficient of \(y^2\)),
- b = -3 (coefficient of \(y\)),
- c = 2 (constant term).
Factoring
Factoring is one method for solving quadratic equations, and it involves expressing the equation as a product of its factors. The factored form of \(y^2 - 3y + 2\) is \((y - 1)(y - 2) = 0\).
This involves finding two numbers that multiply to give the constant term (in this case, 2) and add up to give the linear coefficient (in this case, -3). Here, the factors are -1 and -2 because:
This involves finding two numbers that multiply to give the constant term (in this case, 2) and add up to give the linear coefficient (in this case, -3). Here, the factors are -1 and -2 because:
- -1 * -2 = 2
- -1 + -2 = -3
- \((y - 1) = 0 \Rightarrow y = 1\)
- \((y - 2) = 0 \Rightarrow y = 2\)
Solving for Roots
The roots of an equation are its solutions. Once we have factored the quadratic equation \((y - 1)(y - 2) = 0\), we find the roots by setting each factor equal to zero.
This gives us two simple equations to solve:
This gives us two simple equations to solve:
- \((y - 1) = 0 => y = 1\)
- \((y - 2) = 0 => y = 2\)
- If \(y = 1\): \(x^{1/4} = 1 => x = 1^4 = 1\)
- If \(y = 2\): \(x^{1/4} = 2 => x = 2^4 = 16\)
Raising Exponents
Raising exponents, or exponentiation, is a key mathematical operation used to solve equations involving roots and powers. When we solve for \(x\) after substitution, we often need to 'raise' both sides of the equation to a power to isolate \(x\).
In our steps, after finding that \(y = 1\) or \(y = 2\), we used exponentiation to convert these back to \(x\). For example:
Understanding how to raise exponents correctly ensures you will find the accurate values for solutions.
In our steps, after finding that \(y = 1\) or \(y = 2\), we used exponentiation to convert these back to \(x\). For example:
- When \(y = 1\): \(x^{1/4} = 1\)
Raising both sides to the 4th power, \(x = 1^4\), which simplifies to \(x = 1\). - When \(y = 2\): \(x^{1/4} = 2\)
Raising both sides to the 4th power, \(x = 2^4\), which simplifies to \(x = 16\).
Understanding how to raise exponents correctly ensures you will find the accurate values for solutions.
