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Magazine Chapter 8: Problem 79
Find all real and imaginary solutions to each equation. $$a^{-2}-2 a^{-1}+5=0$$
Short Answer
Expert verified
The solutions are \( a = \frac{1}{5} - \frac{2i}{5}\) and \( a = \frac{1}{5} + \frac{2i}{5} \).
Step by step solution
01
Convert the equation
Rewrite the equation using a substitution method to simplify it. Let鈥檚 set \(u = a^{-1}\), which means \(u = \frac{1}{a}\). Therefore, \(a^{-2}\) will be \(u^2\). The equation now becomes: \[ u^2 - 2u + 5 = 0 \]
02
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:
03
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.
04
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 \(u = 1 - 2i\).
05
Convert back to \(a\)
Since \(u = \frac{1}{a}\), we take the reciprocals of the solutions for \(u\): \[ a = \frac{1}{1 + 2i} \] and \[ a = \frac{1}{1 - 2i} \]
06
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, \[ \frac{1}{1- 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{1 + 2i}{1 - (2i)^2} = \frac{1 + 2i}{1 + 4} = \frac{1 + 2i}{5} = \frac{1}{5} + \frac{2i}{5} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
Understanding quadratic equations is crucial for solving a broad range of mathematical problems. A quadratic equation takes the form:
- \[ ax^2 + bx + c = 0 \]
- \[ u^2 - 2u + 5 = 0 \]
- \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
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:
- \[ \Delta = b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16 \]
- \[ u = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i \]
Discriminant Calculation
The discriminant in a quadratic equation determines the nature of its roots. It is calculated as:
- \[ \Delta = b^2 - 4ac \]
- 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.
- \[ \Delta = -16 \]
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:
- \[ \frac{1}{1 + 2i} \times \frac{1 - 2i}{1 - 2i} = \frac{1 - 2i}{(1 + 2i)(1 - 2i)} = \frac{1 - 2i}{1 + 4} = \frac{1 - 2i}{5} = \frac{1}{5} - \frac{2i}{5} \]
- \[ \frac{1}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{1 + 2i}{1 + 4} = \frac{1 + 2i}{5} = \frac{1}{5} + \frac{2i}{5} \]
