Problem 65 - Precalculus Mathematics for Calculus

Question

Find all solutions of the equation and express them in the form \(a+b i\) $$2 x^{2}-2 x+1=0$$

Step-by-Step Solution

Answer

The solutions are \( x_1 = \frac{1}{2} + \frac{1}{2}i \) and \( x_2 = \frac{1}{2} - \frac{1}{2}i \).

Step 1: Identify the form and coefficients

The given equation is a quadratic equation of the form \( ax^2 + bx + c = 0 \). Here, \( a = 2 \), \( b = -2 \), and \( c = 1 \).

Step 2: Calculate the discriminant

The discriminant \( \Delta \) of a quadratic equation is given by \( \Delta = b^2 - 4ac \). Substitute the values: \( \Delta = (-2)^2 - 4 \cdot 2 \cdot 1 = 4 - 8 = -4 \). The negative discriminant indicates complex solutions.

Step 3: Use the quadratic formula

The quadratic formula is given by \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \). Substitute the known values:\( x = \frac{2 \pm \sqrt{-4}}{4} \).

Step 4: Simplify the square root of the discriminant

Since \( \Delta = -4 \), we have \( \sqrt{-4} = 2i \). Substitute into the quadratic formula: \( x = \frac{2 \pm 2i}{4} \).

Step 5: Simplify to find solutions

Split the expression into real and imaginary parts: \( x = \frac{2}{4} \pm \frac{2i}{4} \). Simplify to obtain \( x = \frac{1}{2} \pm \frac{1}{2}i \). The solutions can be expressed in the form \( a + bi \) as \( x_1 = \frac{1}{2} + \frac{1}{2}i \) and \( x_2 = \frac{1}{2} - \frac{1}{2}i \).

Key Concepts

Understanding Complex Numbers

Complex numbers arise when we deal with the square roots of negative numbers. Typically, they are expressed in the form \(a+bi\), where \(a\) is the real part and \(bi\) is the imaginary part. Here, \(b\) is a real number, and \(i\) is the imaginary unit, defined by \(i^2 = -1\).

• Every complex number has a real part and an imaginary part.
• The imaginary unit \(i\) allows us to explore solutions beyond real numbers, where we cannot take real square roots of negative numbers.

When dealing with quadratic equations with negative discriminants, complex numbers become essential. They ensure that every quadratic equation can have two solutions, even if they are not real solutions. In this exercise, we found the roots \(x = \frac{1}{2} \pm \frac{1}{2}i\), neatly fitting into the \(a + bi\) form, showing both a real and an imaginary part.

Applying the Quadratic Formula

The quadratic formula is a powerful tool to find solutions to any quadratic equation of the form \(ax^2 + bx + c = 0\). It is expressed as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• This formula directly computes the roots by substituting the coefficients \(a\), \(b\), and \(c\).
• It helps determine whether the solutions are real or complex based on the discriminant \(b^2 - 4ac\).

In the exercise, when substituting the values \(a = 2\), \(b = -2\), and \(c = 1\), and applying the quadratic formula, we discover complex solutions. This underscores the importance of using this formula correctly to handle both real and complex roots.

Calculating the Discriminant

The discriminant is a key component in determining the nature of the roots of a quadratic equation.
It is calculated using the expression \( b^2 - 4ac \).

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If it is zero, there is exactly one real solution.
• If the discriminant is negative, as in our original problem, this indicates that the solutions are complex and not real.

In this specific exercise, the discriminant calculated as \(\Delta = -4\) signifies complex solutions. The negative sign tells us directly that square roots will involve imaginary numbers, leading to solutions in the form \( a + bi \). This clear distinction provides insight into the nature of solutions without solving the entire equation initially.