Problem 92 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$625 x^{-4}-125 x^{-2}+4=0$$

Step-by-Step Solution

Answer

The solutions are \( \pm 2.5 \) and \( \pm 5 \).

Step 1 - Substitute Variable

Let’s simplify the equation by substituting a new variable. Let’s set \(y = x^{-2}\). Consequently, \(y^{2} = (x^{-2})^{2} = x^{-4}\). Substituting these into the original equation results in: \[ 625 y^{2} - 125 y + 4 = 0 \]

Step 2 - Identify Quadratic Equation

Now, notice that \(625 y^{2} - 125 y + 4 = 0\) is a quadratic equation in the variable \(y\). A general quadratic equation has the form \(ay^{2} + by + c = 0\). For our specific equation, \(a = 625\), \(b = -125\), and \(c = 4\).

Step 3 - Use the Quadratic Formula

We can solve the quadratic equation using the quadratic formula: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting in the values for \(a\), \(b\), and \(c\): \[ y = \frac{125 \pm \sqrt{(-125)^2 - 4 \cdot 625 \cdot 4}}{2 \cdot 625} \]

Step 4 - Calculate Discriminant

Calculate the value inside the square root (the discriminant): \[ (-125)^2 - 4 \cdot 625 \cdot 4 = 15625 - 10000 = 5625 \] This means the equation now looks like: \[ y = \frac{125 \pm \sqrt{5625}}{1250} \]

Step 5 - Simplify the Square Root

We take the square root of \(5625\): \( \sqrt{5625} = 75 \) So, \( y = \frac{125 \pm 75}{1250} \) which simplifies to two solutions: \[ y = \frac{200}{1250} = \frac{2}{12.5} = 0.16 \] and \[ y = \frac{50}{1250} = 0.04 \]

Step 6 - Back Substitute for x

Recall that \(y = x^{-2}\), so: For \( y = 0.16 \): \[ 0.16 = x^{-2} \] Taking the reciprocal of both sides, \(x^{2} = \frac{1}{0.16} = 6.25\) Thus, \[ x = \pm \sqrt{6.25} = \pm 2.5 \] Similarly, for \( y = 0.04 \): \[ 0.04 = x^{-2} \] Taking the reciprocal: \[ x^2 = 25 \] Thus, \[ x = \pm \sqrt{25} = \pm 5 \]

Key Concepts

Quadratic Substitution

To solve the given equation, it helps to perform a quadratic substitution. The original equation is: \(625 x^{-4} - 125 x^{-2} + 4 = 0\). Complex equations like this can be simplified by recognizing patterns. Here, set a new variable to ease the complexity. Let's use \(y = x^{-2}\). Therefore, \(y^{2} = (x^{-2})^{2} = x^{-4}\). Now, substitute \(y\) and \(y^2\) into the equation: \(625 y^{2} - 125 y + 4 = 0\). This transforms the original equation into a standard quadratic form.

Discriminant

To solve the quadratic equation, we use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, the values of \(a\), \(b\), and \(c\) are \(625\), \(-125\), and \(4\), respectively. A critical part of solving this is calculating the discriminant \(\Delta\), which is inside the square root: \(\Delta = b^2 - 4ac\). For our specific equation, \((-125)^2 - 4 \cdot 625 \cdot 4 = 15625 - 10000 = 5625\). The discriminant tells us the nature of the roots. Since \(\Delta = 5625\) is a positive perfect square, our quadratic equation has two distinct real roots.

Back Substitution

After solving for \(y\), we need to find \(x\) by back substitution. From the quadratic formula, we found two solutions for \(y\): \(y = 0.16\) and \(y = 0.04\). Recall that \(y = x^{-2}\). For \(y = 0.16\): \(0.16 = x^{-2}\), which translates to \((x^{-2})^{-1} = x^{2}\), yielding: \(x^{2} = \frac{1}{0.16} = 6.25\). Taking the square root, \(x = \pm \sqrt{6.25} = \pm 2.5\). Likewise, for \(y = 0.04\): \(0.04 = x^{-2}\), then, \(x^{2} = \frac{1}{0.04} = 25\). Taking the square root, \(x = \pm \sqrt{25} = \pm 5\). Thus, the final solutions are \(x = \pm 2.5\) and \(x = \pm 5\).