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Chapter 3: Problem 39

Solve. $$\sqrt[3]{5 x-2}=-3$$

Short Answer

Expert verified
x = -5

Step by step solution

01

Isolate the radical expression

The given equation is \(\root {3} \relax {5x-2} = -3\). The radical expression \(\root {3} \relax {5x-2}\) is already isolated.
02

Remove the cube root by cubing both sides

To eliminate the cube root, cube both sides of the equation: \(\big( \root {3} \relax {5x-2} \big) ^{3} = (-3)^{3}\). This simplifies to: \(5x - 2 = -27\).
03

Solve for x

To isolate \(x\), add 2 to both sides of the equation: \(5x - 2 + 2 = -27 + 2\), which simplifies to: \(5x = -25\). Finally, divide both sides by 5: \(x = -5\).

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Key Concepts

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

Understanding Cube Roots
When solving cubic equations, it's often necessary to deal with cube roots. In our example, the expression \(\root {3} \relax {5x-2} \) represents the cube root of \( 5x-2 \). A cube root of a number \( y \) is a value that, when raised to the third power, gives \( y \). For instance, the cube root of 27 is 3 because \( 3^3 = 27 \). Cube roots help us find values that, when multiplied by themselves three times, return the original number. In equations, removing the cube root can simplify expressions and make it easier to solve for variables.
Isolating Variables in Equations
In algebra, isolating the variable means getting the variable alone on one side of the equation. This process makes it easier to solve for the variable. Let's look at our example again. The equation is \( \root {3} \relax {5x-2} = -3 \). Initially, the variable expression \( 5x - 2 \) is within a cube root. Since the cube root term is already isolated, we can proceed by eliminating the cube root. To do this, we cube both sides of the equation: \( \big( \root {3} \relax {5x-2} \big) ^{3} = (-3)^{3} \). This simplifies to \( 5x - 2 = -27 \), isolating a linear variable expression.
Solving Basic Algebraic Equations
Now that we've isolated our variable expression, we can use basic algebra to solve for \( x \). We're left with the equation \( 5x - 2 = -27 \). We need to get \( x \) alone, which first involves moving constants to the opposite side. We do this by adding 2 to both sides: \( 5x - 2 + 2 = -27 + 2 \), leading to \( 5x = -25 \). Next, we solve for \( x \) by dividing both sides by 5: \( 5x/5 = -25/5 \), resulting in \( x = -5 \). These basic algebra steps can be applied to many equations, isolating and solving for variables efficiently.

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