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Chapter 1: Problem 42

Find the real solutions, if any, of each equation. $$ x^{3 / 4}-9 x^{1 / 4}=0 $$

Short Answer

Expert verified
The real solutions are \( x = 0 \) and \( x = 81 \).

Step by step solution

01

Substitute with a new variable

Let \( y = x^{1/4} \). Then the equation \( x^{3/4} - 9x^{1/4} = 0 \) can be written as \( y^3 - 9y = 0 \).
02

Factor the equation

Factor the equation \( y^3 - 9y = 0 \). This gives \( y(y^2 - 9) = 0 \).
03

Solve for y

Set each factor equal to zero: \( y = 0 \) or \( y^2 - 9 = 0 \). Solve \( y^2 - 9 = 0 \) to get \( y^2 = 9 \) so \( y = 3 \) or \( y = -3 \). Thus, \( y = 0, 3, -3 \).
04

Substitute back to solve for x

Recall that \( y = x^{1/4} \). Therefore, we have \( x^{1/4} = 0 \), \( x^{1/4} = 3 \), and \( x^{1/4} = -3 \). Solving these, we get: \( x = 0 \) for \( x^{1/4} = 0 \), \( x = 3^4 = 81 \) for \( x^{1/4} = 3 \), and \( x = (-3)^4 = 81 \) for \( x^{1/4} = -3 \).
05

Verify solutions

The solutions found are \( x = 0 \) and \( x = 81 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Equations
Factoring equations is a method used to simplify and solve algebraic expressions. It involves breaking down a complex equation into simpler, multiply factors. For example, let's look at the expression from our problem: \( y^3 - 9y = 0 \). We can factor out a 'y' from this equation,
which gives us \( y(y^2 - 9) = 0 \).
Factoring can make solving equations much easier, especially polynomial equations.
Remember, factoring is essential for breaking down complex problems into more solvable parts.
Substitution Method
The substitution method involves replacing a variable with another expression.
This is useful for simplifying complex equations. In our original problem, we substituted
\( y = x^{1/4} \). This changes the equation from \( x^{3/4} - 9x^{1/4} = 0 \)
to \( y^3 - 9y = 0 \). Substitution makes the problem easier to solve
by reducing the complexity of the variable's exponent.
Solving Polynomial Equations
Polynomial equations are equations that set a polynomial equal to zero. In the equation
\( y^3 - 9y = 0 \), we first factored it to \( y(y^2 - 9) = 0 \). We then set each factor
equal to zero: \( y = 0 \) and \( y^2 - 9 = 0 \). Solving \( y^2 - 9 = 0 \), we get \( y^2 = 9 \),
which results in \( y = 3 \) or \( y = -3 \). Solving polynomial equations often involves
breaking them down into simpler steps, just as we did here.
Verifying Solutions
Verifying solutions is an important step to ensure the solutions are correct.
Here, after solving for \( y = 0, 3, -3 \), we substituted back to get \( x \). We found that \( x = 0 \)
or \( x = 81 \). Finally, we verified by plugging these values back into the original equation
\( x^{3/4} - 9x^{1/4} = 0 \). Both values satisfy the equation, confirming they are correct.
Verifying helps to catch mistakes and confirm that our solutions are correct.

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Most popular questions from this chapter

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Find the real solutions, if any, of each equation. $$ \left(\frac{y}{y-1}\right)^{2}=\frac{6 y}{y-1}+7 $$

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