Problem 81 - Algebra for College Students

Question

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

Step-by-Step Solution

Answer

The solutions are \(x = 1 + i\) and \(x = 1 - i\).

Step 1: Rewrite the Equation with a Substitution

Let \(u = 2x - 1\). Then, we can rewrite the equation as \(u^2 - 2u + 5 = 0\).

Step 2: Solve the Quadratic Equation

To solve \(u^2 - 2u + 5 = 0\), use the quadratic formula: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c=5\). Plugging in these values, \(u = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(5)}}{2(1)}\).

Step 3: Calculate the Discriminant

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

Step 4: Find the Complex Solutions for u

Substitute the values into the quadratic formula to find u: \(u = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i\). Therefore, \(u = 1 + 2i \text{ or } u = 1 - 2i\).

Step 5: Substitute Back to Solve for x

Recall that \(u = 2x - 1\). Substitute \(u\) with each solution: \(1 + 2i = 2x - 1\) and \(1 - 2i = 2x - 1\).

Step 6: Solve for x

Solve each equation for \(x\): \(2x = 2 + 2i\) and \(2x = 2 - 2i\). Thus, \(x = 1 + i\) and \(x = 1 - i\).

Key Concepts

Complex Solutions

In mathematics, solutions to equations can be either real or complex. Real solutions are numbers that we see on a number line like 1, 2, or -5. Complex solutions, on the other hand, involve the imaginary unit, denoted as i, where i equals the square root of -1. For example, a complex solution might look like 1 + i or 3 - 2i.
When solving quadratic equations, if we end up with a negative number inside the square root (discriminant), this means that our equation has complex solutions. In our example, we faced such a situation and the discriminant was -16. This led us to complex solutions. It's essential to recognize such cases and to know that complex numbers expand the range of possible solutions even beyond real numbers.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form: (ax^2 + bx + c = 0) . It states that the solutions can be found using: (\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]) This formula is very useful because it works for all kinds of quadratic equations, whether they have real or complex solutions. In our specific problem, we started with (u^2 - 2u + 5 = 0) and applied the quadratic formula. By substituting the correct values of a, b, and c into the formula, we found that the solutions for u were (1 + 2i \text{ and } 1 - 2i) .Understanding and using the quadratic formula is crucial for solving any quadratic equations you come across.

Discriminant Analysis

The discriminant is a part of the quadratic formula located under the square root ( (b^2 - 4ac) ). It provides crucial information about the nature of the solutions of a quadratic equation. Here's what different discriminant values mean:

• If the discriminant ( (b^2 - 4ac) ) is positive, there are two distinct real solutions.
• If the discriminant is zero, there is exactly one real solution (also called a repeated or double root).
• If the discriminant is negative, there are no real solutions. Instead, there are two complex solutions.

In our problem example, we calculated the discriminant and found it to be -16 ( (4 - 20 = -16) ). The negative value informed us that the equation has complex solutions. Recognizing and calculating the discriminant is a critical step in solving quadratic equations as it hints at the nature of the possible solutions.