Problem 28 - A Graphical Approach to College Algebra

Question

Solve each equation by hand. Do not use a calculator. $$3 x^{-2}-19 x^{-1}+20=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{5}\) and \(x = \frac{3}{4}\).

Step 1: Substitute a Variable

To simplify the equation, let us substitute a new variable. Let \( y = x^{-1} \). This implies \( y^2 = x^{-2} \). Substitute these into the equation: \(3y^2 - 19y + 20 = 0\).

Step 2: Recognize and Use Quadratic Formula

The equation \(3y^2 - 19y + 20 = 0\) is a quadratic equation in the form \(ay^2 + by + c = 0\). Here, \(a = 3\), \(b = -19\), and \(c = 20\). We can solve it using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 3: Calculate the Discriminant

Find the discriminant (\(b^2 - 4ac\)) to see if real solutions exist. Calculate \((-19)^2 - 4 \cdot 3 \cdot 20 = 361 - 240 = 121\). Since 121 is a positive perfect square, there are two real solutions.

Step 4: Solve for y Using Quadratic Formula

Substitute \(a = 3\), \(b = -19\), and \(c = 20\) into the quadratic formula: \[ y = \frac{19 \pm \sqrt{121}}{6} \]. Calculate \( \sqrt{121} = 11 \). So the solutions are: \[ y = \frac{19 + 11}{6} = 5 \] and \[ y = \frac{19 - 11}{6} = \frac{4}{3} \].

Step 5: Substitute Back to Find x

We earlier let \(y = x^{-1}\). So we have \(x^{-1} = 5\) and \(x^{-1} = \frac{4}{3}\). This implies: \(x = \frac{1}{5}\) or \(x = \frac{3}{4}\).

Step 6: Conclusion

The solutions to the original equation \(3 x^{-2}-19 x^{-1}+20=0\) are \(x = \frac{1}{5}\) and \(x = \frac{3}{4}\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of quadratic equations. These are equations that take the form \( ax^2 + bx + c = 0 \).
If you have numbers for \(a\), \(b\), and \(c\), the quadratic formula gives you the solutions directly:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula provides "real" solutions when the discriminant \(b^2 - 4ac\) is zero or positive. When using the formula, it's important to correctly plug in the values for \( a \), \( b \), and \( c \).
The plus-minus symbol (\( \pm \)) indicates that there are usually two solutions.
These solutions can be the same (when the discriminant is zero), or different (when discriminant is positive). The quadratic formula is applicable to any quadratic equation, making it versatile and essential for solving various mathematical problems.

Discriminant

The discriminant is a specific part of the quadratic formula usually denoted as \( b^2 - 4ac \). It plays a crucial role in determining the nature of the roots of a quadratic equation.
Here's how to interpret the discriminant:

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If it is zero, the equation has exactly one real solution, meaning the roots are repeated.
• Finally, if the discriminant is negative, the equation has no real solutions, but two complex ones.

In our exercise, the discriminant was \(121\), which is a perfect square. This indicated that there were two distinct and real roots.
Calculating the discriminant is usually the first step in solving a quadratic equation as it tells us what to expect.

Variable Substitution

Variable substitution is a technique often used to simplify complex equations. The idea is to replace a troublesome expression with a simpler variable.
This makes the equation easier to manipulate or solve.
In the given exercise, substituting \( y = x^{-1} \) transformed the complex expression into a more straightforward quadratic equation \(3y^2 - 19y + 20 = 0\).
Once the equation is simplified and solved, the final step is to substitute back to find the solution in terms of the original variable.
This method is especially useful for polynomials or radical equations and can significantly simplify computations.