Problem 79 - Algebra for College Students

Question

Find all real and imaginary solutions to each equation. $$a^{-2}-2 a^{-1}+5=0$$

Step-by-Step Solution

Answer

The solutions are \( a = \frac{1}{5} - \frac{2i}{5}\) and \( a = \frac{1}{5} + \frac{2i}{5} \).

Step 1: Convert the equation

Rewrite the equation using a substitution method to simplify it. Let’s set \(u = a^{-1}\), which means \(u = \frac{1}{a}\). Therefore, \(a^{-2}\) will be \(u^2\). The equation now becomes: \[ u^2 - 2u + 5 = 0 \]

Step 2: Apply the quadratic formula

To solve for \(u\), use the quadratic formula: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation \(u^2 - 2u + 5 = 0\), \(a = 1\), \(b = -2\), and \(c = 5\). Substitute these values into the formula:

Step 3: Calculate the discriminant

Calculate the discriminant \(\Delta\): \[ \Delta = b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16 \] Since the discriminant is negative (-16), we will have complex solutions.

Step 4: Find the complex solutions for \(u\)

Given the negative discriminant, the solutions for \(u\) will be: \[ u = \frac{-(-2) \pm \sqrt{-16}}{2(1)} = \frac{2 \pm 4i}{2} = 1 \pm 2i \] So, \(u = 1 + 2i\) and \(u = 1 - 2i\).

Step 5: Convert back to \(a\)

Since \(u = \frac{1}{a}\), we take the reciprocals of the solutions for \(u\): \[ a = \frac{1}{1 + 2i} \] and \[ a = \frac{1}{1 - 2i} \]

Step 6: Rationalize the denominators

To simplify \(\frac{1}{1+2i}\) and \(\frac{1}{1-2i}\), multiply the numerator and denominator by the complex conjugate of the denominator: \[ \frac{1}{1+ 2i} \times \frac{1 - 2i}{1 - 2i} = \frac{1 - 2i}{1^2 - (2i)^2} = \frac{1 - 2i}{1 + 4} = \frac{1 - 2i}{5} = \frac{1}{5} - \frac{2i}{5} \] Similarly, \[ \frac{1}{1- 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{1 + 2i}{1 - (2i)^2} = \frac{1 + 2i}{1 + 4} = \frac{1 + 2i}{5} = \frac{1}{5} + \frac{2i}{5} \]

Key Concepts

Quadratic Equation

Understanding quadratic equations is crucial for solving a broad range of mathematical problems. A quadratic equation takes the form:

• \[ ax^2 + bx + c = 0 \]

The equation provided in the exercise can be simplified to this standard form using substitution. By setting \( u = a^{-1} \), the original equation transforms into:

• \[ u^2 - 2u + 5 = 0 \]

This new equation is now a simple quadratic equation in terms of \( u \).To solve any quadratic equation, you can use the quadratic formula:

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

Here, you plug in the coefficients \( a, b, \) and \( c \) from the equation. For this problem, they are \( a = 1 \), \( b = -2 \), and \( c = 5 \).

Complex Numbers

Complex numbers extend the idea of real numbers by including imaginary components. The imaginary unit \( i \) satisfies \( i^2 = -1 \). When the discriminant of a quadratic equation is negative, it produces complex solutions.In our case, after calculating the discriminant, we get:

• \[ \Delta = b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16 \]

Since \( \Delta = -16 \), the solutions will involve imaginary numbers. Solving for \( u \), we have:

• \[ u = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i \]

Thus, we obtain two complex solutions: \( u = 1 + 2i \) and \( u = 1 - 2i \).

Discriminant Calculation

The discriminant in a quadratic equation determines the nature of its roots. It is calculated as:

• \[ \Delta = b^2 - 4ac \]

Based on its value:

• If \( \Delta > 0 \), there are two distinct real roots.
• If \( \Delta = 0 \), there is one real root.
• If \( \Delta < 0 \), there are two complex roots.

For our quadratic equation \( u^2 - 2u + 5 = 0 \), we calculated the discriminant as:

• \[ \Delta = -16 \]

Since it is negative, the equation has two complex roots, confirming the presence of imaginary numbers.

Rationalization

Rationalization is the process of eliminating the imaginary unit from the denominator of a fraction. To achieve this, we multiply by the complex conjugate.For the solutions \[ a = \frac{1}{1 + 2i} \] and \[ a = \frac{1}{1 - 2i} \], we use their complex conjugates for rationalization. The steps are:

• \[ \frac{1}{1 + 2i} \times \frac{1 - 2i}{1 - 2i} = \frac{1 - 2i}{(1 + 2i)(1 - 2i)} = \frac{1 - 2i}{1 + 4} = \frac{1 - 2i}{5} = \frac{1}{5} - \frac{2i}{5} \]

Similarly, we handle the other solution:

• \[ \frac{1}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{1 + 2i}{1 + 4} = \frac{1 + 2i}{5} = \frac{1}{5} + \frac{2i}{5} \]

These steps ensure our final solutions are expressed in a simple form, making them easier to interpret.