Problem 90 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$2 x^{-2 / 5}-x^{-1 / 5}-1=0$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Substitute with a New Variable

Let us introduce a new variable to simplify the equation. Let \( y = x^{-1/5} \). This substitution gives: \( 2y^2 - y - 1 = 0 \).

Step 2: Solve the Quadratic Equation

We need to solve the quadratic equation \( 2y^2 - y - 1 = 0 \). To do this, use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -1 \), and \( c = -1 \).

Step 3: Apply the Quadratic Formula

Substituting the values into the quadratic formula: \( y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \). This gives two solutions: \( y = 1 \) and \( y = -\frac{1}{2} \).

Step 4: Revert Substitution to Original Variable

Recall that \( y = x^{-1/5} \). Therefore, we have two cases: \( x^{-1/5} = 1 \) and \( x^{-1/5} = -\frac{1}{2} \).

Step 5: Solve for \( x \)

For \( x^{-1/5} = 1 \), raise both sides to the power of -5: \( x = 1^{-5} = 1 \). For \( x^{-1/5} = -\frac{1}{2} \), raising both sides to the power of -5 gives a non-real number since the base is negative and not a principal root, so this part of the solution is discarded.

Key Concepts

Substitution Method

The substitution method simplifies complex equations by introducing a new variable. In our example, we start with the equation:
$$2 x^{-2 / 5}-x^{-1 / 5}-1=0.$$
This is a complicated equation due to the exponents on x. By letting y be equal to one of the terms, we can turn the equation into a more familiar form. Here, we make the substitution:
Let y = x^{-1/5}.
Now, the equation transforms to:
$$2y^2 - y - 1 = 0.$$
This substitution helps to manage the tricky exponents and turns our problem into a standard quadratic equation, which is much easier to solve.

Quadratic Formula

Once the equation is simplified using substitution, we can solve it using the quadratic formula. For a quadratic equation in the form ax^2 + bx + c = 0, the quadratic formula is:
$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
Here, the coefficients are

• a = 2
,
• b = -1
,
• c = -1
.

We substitute these values into the quadratic formula:
$$y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm\sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}.$$ This gives us two solutions for y:

• y = 1
and
• y = -1/2

Now the quadratic equation is solved, and we can return to our original variable.

Real and Non-Real Solutions

Finally, we need to revert back to our original variable and understand the nature of the solutions. Recall the substitution:
y = x^{-1/5}.
Our two solutions for y were 1 and -1/2. We convert back to x for each solution.

For x^{-1/5} = 1, raising both sides to the power of -5 gives us:
$$x = 1^{-5} = 1.$$
For x^{-1/5} = -1/2, raising both sides to the power of -5 would result in a non-real number because we are raising a negative number to a negative fractional power. In this context, such solutions are considered non-real and therefore are discarded.

Real solutions make sense within the original exercise while non-real solutions do not. Understanding this distinction is crucial in algebraic equations.