Problem 60 - Algebra for College Students

Question

Find all real solutions to each equation. $$x^{2 / 3}-x^{1 / 3}-20=0$$

Step-by-Step Solution

Answer

The real solutions are x = 125 and x = -64.

Step 1: Introduce a Substitute Variable

To make the equation easier to solve, let’s introduce a new variable. Define \[ y = x^{1/3} \]. Therefore, \[ y^2 = (x^{1/3})^2 = x^{2/3} \]. This simplifies the original equation to \[ y^2 - y - 20 = 0 \].

Step 2: Solve the Quadratic Equation

Now solve the quadratic equation \[ y^2 - y - 20 = 0 \].Factor the quadratic equation: \[ (y - 5)(y + 4) = 0 \].This gives the solutions \[ y = 5 \] and \[ y = -4 \].

Step 3: Substitute Back the Original Variable

Now substitute back \[ y = x^{1/3} \]:For \[ y = 5 \]: \[ x^{1/3} = 5 \], therefore, \[ x = 5^3 = 125 \].For \[ y = -4 \]: \[ x^{1/3} = -4 \], therefore, \[ x = (-4)^3 = -64 \].

Step 4: Verify the Solutions

Check both solutions in the original equation \[ x^{2/3} - x^{1/3} - 20 = 0 \]: For \[ x = 125 \]: \[ 125^{2/3} - 125^{1/3} - 20 \] simplifies to \[ 25 - 5 - 20 = 0 \] which is true.For \[ x = -64 \]: \[ (-64)^{2/3} - (-64)^{1/3} - 20 \] simplifies to \[ 16 - (-4) - 20 = 0 \] which is also true.

Key Concepts

Substitute Variable

In dealing with radical equations, one of the first steps can be to simplify the equation by introducing a new variable. This makes the equation easier to handle. For example, if we start with the equation \[ x^{2/3} - x^{1/3} - 20 = 0 \], it can be simplified by letting \[ y = x^{1/3} \]. This substitution transforms our original problem into a more familiar form. \
By making the substitution, \[ y^2 = x^{2/3} \], the equation becomes \[ y^2 - y - 20 = 0 \]. Now, instead of working with radicals, we've transformed the problem into solving a quadratic equation—something much more straightforward!

Quadratic Equation

Now that we have simplified our radical equation, we are left with a quadratic equation: \[ y^2 - y - 20 = 0 \]. Quadratic equations take the form of \[ ax^2 + bx + c = 0 \]. In this case, \[ a = 1 \], \[ b = -1 \], and \[ c = -20 \].
To solve this quadratic equation, we can factor it: \[ y^2 - y - 20 = 0 \].
This factors into \[ (y - 5)(y + 4) = 0 \], giving us the solutions \[ y = 5 \] and \[ y = -4 \]. These are the solutions for \[ y \], but remember, our original variable was \[ x \]. So, the next step is to substitute \[ y \] back into our expression involving \[ x \].

Solution Verification

The final and very important step is to verify that our solutions are correct by substituting them back into the original equation. Let's consider both solutions:
For \[ y = 5 \]: We had \[ x^{1/3} = 5 \], hence \[ x = 5^3 = 125 \]. Substituting this into the original equation \[ x^{2/3} - x^{1/3} - 20 = 0 \], we get: \[ 125^{2/3} - 125^{1/3} - 20 \]. Simplified, this becomes: \[ 25 - 5 - 20 = 0 \] which holds true.
For \[ y = -4 \]: We had \[ x^{1/3} = -4 \], hence \[ x = (-4)^3 = -64 \]. Substituting this into the original equation, we get: \[ (-64)^{2/3} - (-64)^{1/3} - 20 = 0 \]. Simplified, this becomes: \[ 16 - (-4) - 20 = 0 \] which also holds true.
Thus, both \[ x=125 \] and \[ x=-64 \] are verified as real solutions to the original radical equation.