Problem 27 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation by hand. Do not use a calculator. $$5 x^{-2}+13 x^{-1}=28$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{5}{7} \) and \( x = -\frac{1}{4} \).

Step 1: Make a Substitution

To simplify the equation, let's make a substitution. Let \( y = x^{-1} \). Then, \( y^2 = x^{-2} \). Using this substitution, our equation becomes \( 5y^2 + 13y = 28 \).

Step 2: Rearrange the Equation

Rearrange the equation to standard quadratic form: \( 5y^2 + 13y - 28 = 0 \). This is now a quadratic equation in terms of \( y \).

Step 3: Solve the Quadratic Equation

Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b = 13 \), and \( c = -28 \). Calculate the discriminant: \( b^2 - 4ac = 13^2 - 4 \times 5 \times (-28) = 169 + 560 = 729 \).

Step 4: Calculate the Roots

The roots are calculated as: \[y = \frac{-13 \pm \sqrt{729}}{10} = \frac{-13 \pm 27}{10}.\]So, \( y_1 = \frac{14}{10} = 1.4 \) and \( y_2 = \frac{-40}{10} = -4 \).

Step 5: Reverse the Substitution

Recall that \( y = x^{-1} \), so \( x = \frac{1}{y} \). Thus, for \( y_1 = 1.4 \), \( x = \frac{1}{1.4} = \frac{5}{7} \), and for \( y_2 = -4 \), \( x = \frac{1}{-4} = -\frac{1}{4} \).

Step 6: Verify Solutions

Substitute \( x = \frac{5}{7} \) and \( x = -\frac{1}{4} \) back into the original equation to verify they are indeed solutions. For \( x = \frac{5}{7} \):\[ 5 \left(\frac{7}{5}\right)^2 + 13 \left(\frac{7}{5}\right) = 28. \]And for \( x = -\frac{1}{4} \):\[ 5 \left(-4\right)^2 + 13 \left(-4\right) = 28. \]Both calculations confirm that these are valid solutions.

Key Concepts

Substitution Method

When faced with complex algebraic equations, the substitution method can be a powerful tool to simplify the problem. In the given exercise, we started by replacing a tricky term to make the equation easier to manage. The equation, initially presented as \(5x^{-2} + 13x^{-1} = 28\), involved challenging negative exponents. By letting \(y = x^{-1}\), we transformed these negative exponents into simple forms:\( y^2 = x^{-2} \), allowing us to rewrite the entire equation as \(5y^2 + 13y = 28\).

This substitution turns a seemingly complex equation into a standard quadratic formula in terms of \(y\). It is a strategic move because quadratic equations are more straightforward to solve using established methods like the quadratic formula. This tactic is especially useful when dealing with equations that contain exponents or other complicating factors. Breaking down the equation into simpler components can provide a clear path to the solution.

Quadratic Formula

The quadratic formula is a dependable tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). In our transformed equation from the substitution method \(5y^2 + 13y - 28 = 0\), the quadratic formula allows us to find the values of \(y\) that satisfy this equation.

The quadratic formula is written as:

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a = 5\), \(b = 13\), and \(c = -28\). We first calculate the discriminant \(b^2 - 4ac\). This value, \(b^2 - 4ac = 729\), is crucial because it determines if real solutions exist. A positive discriminant implies two real solutions. Solving gives us the roots \(y_1 = 1.4\) and \(y_2 = -4\).

Understanding how to apply the quadratic formula is vital for solving quadratic equations effectively, especially in situations where factoring isn't straightforward. The formula provides a universal method to obtain the roots that satisfy any quadratic equation.

Reversing Substitution

Reversing the substitution is the final step in solving the original equation. After finding the values of \(y\), we must convert them back to the original variable \(x\) to complete the solution. Since we substituted \(y = x^{-1}\), reversing this step requires us to express \(x\) in terms of \(y\).

We do this by rearranging the substitution equation: \(x = \frac{1}{y}\). Thus, plug in the values of \(y\) we found earlier:

• For \(y_1 = 1.4\), \(x = \frac{1}{1.4} = \frac{5}{7}\).
• For \(y_2 = -4\), \(x = \frac{1}{-4} = -\frac{1}{4}\).

This step is crucial because it translates the solution from our simplified world of \(y\) back to the original context in terms of \(x\). It's essential to verify these values in the original equation to ensure they work, confirming the solution's correctness.