Problem 72 - Beginning and Intermediate Algebra

Question

The following exercises are not grouped by type. Solve each equation. (Exercises 83 and 84 require knowledge of complex numbers.) \(\left(x-\frac{1}{2}\right)^{2}+5\left(x-\frac{1}{2}\right)-4=0\)

Step-by-Step Solution

Answer

The solutions are \( x = \frac{-4 + \sqrt{41}}{2} \) and \( x = \frac{-4 - \sqrt{41}}{2} \).

Step 1: Substitute the variable

Let us substitute the variable to simplify the equation. Let \( y = x - \frac{1}{2} \). This gives us the equation \( y^2 + 5y - 4 = 0 \).

Step 2: Identify coefficients

Identify the coefficients from the quadratic equation \( y^2 + 5y - 4 = 0 \). Here, \( a = 1 \), \( b = 5 \), and \( c = -4 \).

Step 3: Apply the quadratic formula

Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plug in the values: \( y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \).

Step 4: Simplify the quadratic formula

Calculate the discriminant: \( 5^2 - 4 \cdot 1 \cdot (-4) = 25 + 16 = 41 \). Then plug it back into the formula: \( y = \frac{-5 \pm \sqrt{41}}{2} \).

Step 5: Find the roots

Calculate the two possible solutions for \( y \): \( y = \frac{-5 + \sqrt{41}}{2} \) and \( y = \frac{-5 - \sqrt{41}}{2} \).

Step 6: Substitute back to \( x \)

Recall that \( y = x - \frac{1}{2} \). So, \( x - \frac{1}{2} = \frac{-5 + \sqrt{41}}{2} \) and \( x - \frac{1}{2} = \frac{-5 - \sqrt{41}}{2} \).

Step 7: Solve for \( x \)

Solving for \( x \), we get: \( x = \frac{-5 + \sqrt{41}}{2} + \frac{1}{2} = \frac{-4 + \sqrt{41}}{2} \) and \( x = \frac{-5 - \sqrt{41}}{2} + \frac{1}{2} = \frac{-4 - \sqrt{41}}{2} \).

Key Concepts

Complex Numbers

Complex numbers are an extension of real numbers and are used to solve equations that don't have real solutions. A complex number is in the form of \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit. The imaginary unit \(i\) is defined by the property that \(i^2 = -1\).
If the discriminant of a quadratic equation is negative, the solutions will involve complex numbers. For instance, \(-4\) is a discriminant that would be simplified using complex numbers, like \(2 + 3i\) and \(2 - 3i\).
In our exercise, since the discriminant \(41\) is positive, we conclude the solutions are real, indicating we don’t require complex numbers here.

Quadratic Formula

The quadratic formula is a universal method to find the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\).
The formula is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
To use the quadratic formula, you need to identify the coefficients \(a\), \(b\), and \(c\) from the equation. For example, in our equation, substituting \(a = 1\), \(b = 5\), and \(c = -4\), we substitute these into our formula and solve:
\( x = \frac{-5 \pm \sqrt{41}}{2} \). This calculation gives us two possible roots for \(x\).

Discriminant

The discriminant is a part of the quadratic formula and plays a crucial role in determining the nature of the roots of a quadratic equation.
It is denoted by \( \Delta \) and calculated as:
\( \Delta = b^2 - 4ac \)
The value of the discriminant tells us:

• If \( \Delta > 0 \): Two distinct real roots exist.
• If \( \Delta = 0 \): One real double root exists (both roots are the same).
• If \( \Delta < 0 \): Two complex roots exist.

For the given exercise, \( \Delta = 41 \), which is greater than zero. Thus, the quadratic equation has two distinct real roots.

Substitution Method

The substitution method can simplify solving a complex equation by replacing a portion of the equation with a single variable.
This method makes the quadratic equation easier to handle. For instance, the original exercise starts with:
\( \left(x - \frac{1}{2}\right)^{2} + 5\left(x - \frac{1}{2}\right) - 4 = 0 \)
We let \( y = x - \frac{1}{2}\) to simplify the equation to \( y^2 + 5y - 4 = 0 \).
This transformation facilitates our use of the quadratic formula. Once we find \( y \), we substitute back to solve for the original variable \( x \).
Thus, substitution turns a complex problem into a manageable one.