Question

Find the real solutions of each equation. $$ x^{1 / 2}-3 x^{1 / 4}+2=0 $$

Short answer

The real solutions are \( x = 1 \) and \( x = 16 \).

Step-by-step solution

Substitute to simplify

Let’s use the substitution method to simplify the equation. Let’s set \( y = x^{1/4} \). Therefore, \( y^2 = x^{1/2} \). Substituting these into the equation gives: \[ y^2 - 3y + 2 = 0 \]

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: \[ (y - 1)(y - 2) = 0 \]

Solve for y

Set each factor to zero and solve for \( y \): \[ y - 1 = 0 \, \Rightarrow \, y = 1 \] \[ y - 2 = 0 \, \Rightarrow \, y = 2 \]

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, \[ x = 16 \]

State the solutions

The solutions to the original equation are: \( x = 1 \) and \( x = 16 \)

Key concepts

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:

• Original Equation: \(x^{1/2} - 3x^{1/4} + 2 = 0\)
• After Substitution: \(y^2 - 3y + 2 = 0\).

This step can dramatically reduce the difficulty of solving the problem, making the remaining steps much more manageable.

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:

• a = 1 (coefficient of \(y^2\)),
• b = -3 (coefficient of \(y\)),
• c = 2 (constant term).

Recognizing these elements makes it easier to proceed to the next steps.

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:

• -1 * -2 = 2
• -1 + -2 = -3

Once factored, the quadratic equation becomes much simpler to solve because you can set each factor equal to zero and solve for \(y\):

• \((y - 1) = 0 \Rightarrow y = 1\)
• \((y - 2) = 0 \Rightarrow y = 2\)

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:

• \((y - 1) = 0 => y = 1\)
• \((y - 2) = 0 => y = 2\)

These are the values for \(y\) but remember, we initially set \(y = x^{1/4}\). So, we need to substitute back \(y\) values to get the solutions for \(x\). Finally:

• If \(y = 1\): \(x^{1/4} = 1 => x = 1^4 = 1\)
• If \(y = 2\): \(x^{1/4} = 2 => x = 2^4 = 16\)

The roots or solutions to the original equation \(x^{1/2} - 3x^{1/4} + 2 = 0\) are \(x = 1\) and \(x = 16\).

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:

• 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 \(x = 16\).

This technique is useful whenever you need to remove a root or transform an exponent in an equation.
Understanding how to raise exponents correctly ensures you will find the accurate values for solutions.