Problem 90 - Algebra for College Students

Question

Find all real or imaginary solutions to each equation. Use the method of your choice. $$(2 x-1)^{2}=-4$$

Step-by-Step Solution

Answer

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

Step 1: Isolate the Term with the Variable

Start by rewriting the equation to make it easier to solve:\( (2x - 1)^2 = -4 \).

Step 2: Recognize the Nature of the Equation

Notice that the right-hand side of the equation is negative. Since the square of a real number is always non-negative, this implies that we are dealing with imaginary solutions.

Step 3: Use the Definition of Imaginary Numbers

To solve for imaginary numbers, rewrite -4 as 4 multiplied by -1:\[ (2x - 1)^2 = 4(-1) \].This simplifies to:\[ (2x - 1)^2 = 4i^2 \],where \(i\) is the imaginary unit and \(i^2 = -1\).

Step 4: Take the Square Root of Both Sides

Apply the square root to both sides of the equation:\[ \pm\sqrt{(2x - 1)^2} = \pm\sqrt{4i^2} \].This simplifies to:\[ 2x - 1 = \pm2i \].

Step 5: Solve for x

Finally, isolate \(x\) by rearranging the equation:\[ 2x - 1 = 2i \] and \[ 2x - 1 = -2i \].For the first equation:\[ 2x = 2i + 1 \]\[ x = i + \frac{1}{2} \].For the second equation:\[ 2x = -2i + 1 \]\[ x = -i + \frac{1}{2} \].

Key Concepts

Imaginary Numbers

Imaginary numbers might seem tricky, but they're just numbers that, when squared, give a negative result.
Normally, squaring any real number always gives a positive result. That's where imaginary numbers step in.
We symbolize the basic imaginary unit as 'i', where:
* i² = -1
Using imaginary numbers helps us solve equations that don't have real number solutions, especially quadratic equations with negative squares. In our example, \[(2x - 1)^2 = -4\], the equation suggests the solution involves imaginary numbers because no squared real number equals -4.

Square Root

Taking the square root of a number means finding a value that, when multiplied by itself, gives the original number. For example, the square root of 4 is 2, because 2 * 2 = 4.
However, when taking the square root of a negative number, we use imaginary numbers. In our problem, here's how it works:
We need the square root of -4. This can be expressed as the product of the square root of 4 and the square root of -1: \[\text{√}(-4) = \text{√}(4) \times \text{√}(-1)\].
Because \[\text{√}(4)\] is 2 and \[\text{√}(-1)\] is 'i', we get \[\text{±2i}\].
Therefore, for the equation \[\text{√}((2x - 1)^2) = \text{±√}(4i^2)\], we simplify it to \[(2x - 1) = \text{±2i}\].

Isolating the Variable

Isolating the variable means rearranging the equation so that we solve for \[x\].
Let's break it down step-by-step for our problem:
1. We start with \[(2x - 1) = \text{±2i}\].
2. To get rid of -1, we add 1 to both sides of the equation. This gives:
\[2x = 2i + 1\] or \[2x = -2i + 1\].
3. Next, divide by 2 to isolate \[x\]:
For \[(2x = 2i + 1)\], we get \[x = i + \frac{1}{2}\].
For \[(2x = -2i + 1)\], we get \[x = -i + \frac{1}{2}\].
Now, we have successfully isolated \[x\] in both equations.

Complex Solutions

Complex solutions occur when the solutions to an equation have both real and imaginary parts.
These solutions take the form \[a + bi\] where \[a\] and \[b\] are real numbers, and \[i\] is the imaginary unit.
In our example, the solutions we found for \[x\] are:
* \[x = i + \frac{1}{2}\]
* \[x = -i + \frac{1}{2}\]
Both solutions consist of a real part (\[\frac{1}{2}\]) and an imaginary part (\[i\] and \[-i\]).
Therefore, the solutions are complex numbers, which means they are not purely real or imaginary but a combination of both.