Problem 34 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ \sqrt{x}-3 \sqrt[4]{x}-4=0 $$

Step-by-Step Solution

Answer

The real solution is \( x = 256 \).

Step 1: Set Variables

Introduce a variable substitution to simplify the equation. Let \( y = \sqrt[4]{x} \). Then, \( \sqrt{x} = y^2 \). The equation becomes \( y^2 - 3y - 4 = 0 \).

Step 2: Factor the Quadratic

The rewritten equation \( y^2 - 3y - 4 = 0 \) can be factored as \((y - 4)(y + 1) = 0\). This gives two solutions: \( y - 4 = 0 \) and \( y + 1 = 0 \).

Step 3: Solve for y

From the factors, solve for \( y \). The solutions are \( y = 4 \) and \( y = -1 \).

Step 4: Solve for x

Recall \( y = \sqrt[4]{x} \), so \( x = y^4 \). Substitute the solutions for \( y \):- For \( y = 4 \), \( x = 4^4 = 256 \).- \( y = -1 \) is not valid since \( \sqrt[4]{x} \) must be non-negative.

Step 5: Verify Solutions

Substitute \( x = 256 \) back into the original equation to verify:\( \sqrt{256} - 3\sqrt[4]{256} - 4 = 16 - 3(4) - 4 = 16 - 12 - 4 = 0 \).Thus, \( x = 256 \) is valid.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest power of the variable is squared, typically in the form of \( ax^2 + bx + c = 0 \). These equations are fundamental in algebra and can have two solutions. Some key characteristics include:

• The equation is shaped like a parabola when graphed.
• It can have either two real solutions, one real solution, or no real solution.
• Methods to solve them include factorization, completing the square, and using the quadratic formula.

In the given exercise, after substituting \( y = \sqrt[4]{x} \), we transformed the equation into a more familiar quadratic form: \( y^2 - 3y - 4 = 0 \). This adaptation helps us to apply the technique of factoring to find the solutions for \( y \). Let's explore how substitution aids in simplifying such problems.

Variable Substitution

Variable substitution is a powerful method often used to simplify complex equations. By introducing a new variable, we transform the equation into a more tractable form. Here’s how it works in the given exercise:

• We let \( y = \sqrt[4]{x} \), recognizing that \( \sqrt{x} = y^2 \).
• This change converts the original equation into \( y^2 - 3y - 4 = 0 \).

This transformation allows us to focus on solving a quadratic equation, as it's easier to work with compared to the original form. It's important to remember that while variable substitution simplifies the equation, we'll eventually need to substitute back to find the values of the original variable \( x \). This step ensures that the solution meets the conditions of the original problem, particularly regarding valid values for square roots and real numbers.

Real Solutions

In mathematics, real solutions refer to answers to an equation that can be represented as real numbers. Real numbers include all rational and irrational numbers, essentially anything that can be placed on a continuous number line. In the context of our quadratic equation:

• Real solutions are those values that satisfy the equation without leading to non-real or imaginary numbers.
• The exercise highlights the importance of remembering the domain restrictions of functions like square roots.

For instance, the solution \( y = -1 \) was discarded because \( \sqrt[4]{x} \) must be non-negative when dealing with real numbers. Thus, for the valid result \( y = 4 \), the corresponding \( x \) value was verified as \( x = 256 \) by substituting back and checking it in the original equation.
Understanding real solutions ensures that we find meaningful answers that align with the domain the equation is defined over, leading us to correct and applicable results in real-world contexts.