Problem 67
Question
\(47-70\) The given equation involves a power of the variable. Find all real solutions of the equation. $$ \sqrt[3]{x}=5 $$
Step-by-Step Solution
Verified Answer
The solution is \( x = 125 \).
1Step 1: Understand the equation
The equation given is \( \sqrt[3]{x} = 5 \). This means we are looking for a number \( x \) such that when you take its cube root, the result is 5.
2Step 2: Isolate the variable by removing the cube root
To solve for \( x \), we need to remove the cube root. This is done by cubing both sides of the equation. When we cube \( \sqrt[3]{x} \), we get \( x \), and cubing 5 gives \( 5^3 \).
3Step 3: Perform the operation
Now, we carry out the cubing operation: \( x = 5^3 \). Calculate \( 5^3 = 5 \times 5 \times 5 = 125 \). Therefore, \( x = 125 \).
Key Concepts
Cube RootsReal SolutionsIsolation of Variables
Cube Roots
A cube root of a number is a value that, when multiplied by itself three times, gives the original number. In simpler terms, if you have a number, say 27, the cube root would be 3 because \(3 \times 3 \times 3 = 27\). Understanding cube roots is crucial since they help simplify equations involving powers and roots. The notation \( \sqrt[3]{x} \) represents the cube root of \( x \). To "undo" a cube root, you would generally "cube" or raise a number to the third power. When solving equations, identifying when a cube root is involved allows you to correctly manipulate the equation to isolate the variable. Cube roots can also be negative, unlike square roots, which adds an additional layer of complexity to some equations. However, in the problem mentioned, the cube root operation is straightforward as it involves a positive number. This simplicity is why cube roots are often chosen deliberately to filter out complexities when learning the basics of root operations.
Real Solutions
When solving equations, especially those involving cubic terms, it's essential to focus on finding "real solutions." Real solutions are solutions that are not imaginary and can best represent values we can see and quantify in the real world. In the context of the equation \( \sqrt[3]{x} = 5 \), we are seeking the real value of \( x \) that satisfies this equation. We've found that cube roots can yield real numbers easily since cubing preserves realness, unlike square roots which might encounter complex numbers. Real solutions are particularly significant because they often correspond to practical interpretations of problems, such as measurements or counts. Understanding when an equation might yield a real solution is a key skill that helps refine strategies for problem-solving and also offers a clearer, easier path to follow when handling these equations.
Isolation of Variables
Isolation of variables is a key strategy in solving equations. It means manipulating the equation so that the unknown variable stands alone on one side of the equation, and everything else is on the other side. This allows you to find the value of the variable directly. In the problem \( \sqrt[3]{x} = 5 \), we isolate \( x \) by eliminating the cube root from \( x \). This is done by cubing both sides, as cubing is the inverse operation of taking a cube root. Thus, cubing \( \sqrt[3]{x} \) results in \( x \), and cubing 5 results in 125.This strategy ensures that you can handle more complex equations in a systematic way, by repeatedly isolating variables step by step. Keeping the isolation process clear and methodical helps in avoiding mistakes and makes it easier to follow the logical steps to reach the solution.
Other exercises in this chapter
Problem 67
65–68 ? Solve the equation for the variable x. The constants a and b represent positive real numbers. $$ \sqrt{x+a}+\sqrt{x-a}=\sqrt{2} \sqrt{x+6} $$
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Use the discriminant to determine the number of real solutions of the equation. Do not solve the equation. \(x^{2}+r x-s=0 \quad(s>0)\)
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\(67-70=\) Determine the values of the variable for which the expression is defined as a real number. $$ \sqrt{3 x^{2}-5 x+2} $$
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Recall that the symbol \(\overline{z}\) represents the complex con- jugate of \(z .\) If \(z=a+b i\) and \(w=c+d i,\) prove each statement. $$ \overline{z w}=\o
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