Problem 51 - College Algebra Essentials

Question

By making an appropriate substitution. $$ x^{\frac{3}{2}}-2 x^{\frac{3}{4}}+1=0 $$

Step-by-Step Solution

Answer

The solution to the equation \(x^{\frac{3}{2}}-2x^{\frac{3}{4}}+1=0\) is \(x = 1\).

Step 1: Introduce the Substitution

Introduce the substitution \(y=x^{\frac{3}{4}}\). Then, \(y^2 = (x^{\frac{3}{4}})^2 = x^{\frac{3}{2}}\). Substituting this into the equation yields a quadratic equation in terms of y: \(y^2-2y+1=0\).

Step 2: Solve the Quadratic Equation

Solve the quadratic equation through factoring: \((y-1)^2 = 0\). The solution is \(y = 1\).

Step 3: Back Substitute y into Terms of x

Replace y with \(x^{\frac{3}{4}}\) in y=1 to find the positive real solution/s for x: \(x^{\frac{3}{4}} = 1\). Therefore, \(x = (1)^{\frac{4}{3}} = 1\).

Key Concepts

Substitution Method

The substitution method is a powerful technique in algebra that helps us simplify complex equations by replacing variables with a substitute one. This simplification often transforms the equation into a more familiar or manageable form.

In the context of the given exercise, we have the equation: \( x^{\frac{3}{2}}-2x^{\frac{3}{4}}+1=0 \). By introducing a substitution, we aim to transform this expression into an easier-to-solve quadratic equation.

Here, we make the substitution \( y = x^{\frac{3}{4}} \). By substituting, \( y^2 = (x^{\frac{3}{4}})^2 = x^{\frac{3}{2}} \). As a result, our equation becomes \( y^2 - 2y + 1 = 0 \), which is now a quadratic equation in terms of \( y \). This substitution reduces the complexity and makes the problem more approachable.

Factoring Quadratics

Once we have our quadratic equation \( y^2 - 2y + 1 = 0 \), the next step is to solve it by factoring. Quadratic equations are often solved by this method because factoring can reveal the roots or solutions of the equation without further complication.

To factor the equation \( y^2 - 2y + 1 \), we notice that it can be written as \((y-1)^2 = 0\).

• This reveals a "perfect square" trinomial, meaning each factor is the same: \( (y-1) \).
• By setting \( (y-1)^2 = 0 \), we find that \( y = 1 \).
• Factoring wields simplicity, because instead of working with a quadratic, we joyfully deal with a simple linear equation.

Exponents and Powers

Exponents and powers are foundational concepts in algebra and help us express repeated multiplication concisely. In this exercise, understanding how exponents work is crucial to solving the original problem and utilizing substitution effectively.

Here are some key points about exponents relevant to this exercise:

• \( x^{\frac{3}{2}} \) implies \( (x^{\frac{1}{2}})^3 \) or the cube of the square root of \( x \).
• Whereas \( x^{\frac{3}{4}} \) is the fourth root of \( x^3 \).
• The laws of exponents allow transformations into simpler forms, such as \( (x^{\frac{3}{4}})^2 = x^{\frac{3}{2}} \), which substantiates our substitution process.

Understanding the behavior of exponents facilitates the restructuring of equations to apply the substitution method effectively.

Solving Equations

The ultimate goal in algebra is solving equations, meaning finding the value of the variables that make the equation true.

Once we've simplified and factored our quadratic to find \( y = 1 \), our task isn't complete until we express it back in terms of the original variable \( x \).

The final step is to "back substitute" the expression \( y = x^{\frac{3}{4}} \) into our solution of \( y = 1 \). Solving for \( x \), we get:

• Set up the equation from the substitution \( x^{\frac{3}{4}} = 1 \).
• To solve for \( x \), we raise both sides to the reciprocal of \( \frac{3}{4} \), which is \( \frac{4}{3} \).
• This results in \( x = (1)^{\frac{4}{3}} = 1 \).

Thus, the solution \( x = 1 \) satisfies the original equation, completing the solving process.