Problem 87 - College Algebra with Modeling and Visualization

Question

Complete the following. (a) Write the equation as \(a x^{2}+b x+c=0\) with \(a>0\) (b) Calculate the discriminant \(b^{2}-4 a c\) and determine the number of real solutions. (c) Solve the equation. $$ x^{2}+1=x $$

Step-by-Step Solution

Answer

The equation lacks real solutions; its solutions are complex: \(x = \frac{1 \pm i\sqrt{3}}{2}\).

Step 1: Rearrange the Equation

Start with the given equation: \(x^2 + 1 = x\). Move \(x\) to the left side to make the equation equal to zero: \(x^2 - x + 1 = 0\). This is now in the form \(a x^{2}+b x+c=0\), where \(a = 1\), \(b = -1\), and \(c = 1\).

Step 2: Calculate the Discriminant

The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(b^2 - 4ac\). Substitute \(a = 1\), \(b = -1\), and \(c = 1\) into the formula: \((-1)^2 - 4(1)(1) = 1 - 4 = -3\).

Step 3: Determine the Number of Real Solutions

Since the discriminant \(b^2 - 4ac\) is negative (-3), the quadratic equation does not have any real solutions. Instead, it has two complex solutions.

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = 1\), \(b = -1\), and \(c = 1\) into the formula: \(x = \frac{-(-1) \pm \sqrt{-3}}{2 \times 1} = \frac{1 \pm \sqrt{-3}}{2}\). Simplify further: \(x = \frac{1 \pm i\sqrt{3}}{2}\), where \(i\) is the imaginary unit.

Key Concepts

Discriminant

In quadratic equations, the discriminant is a key component that helps us understand the nature and number of solutions. It is calculated using the formula \(b^2 - 4ac\), derived from the general form of a quadratic equation, \(ax^2 + bx + c = 0\).

• When the discriminant is positive, the quadratic equation has two distinct real solutions.
• If the discriminant is zero, there is exactly one real solution, which is a repeated root.
• A negative discriminant indicates no real solutions but implies two complex solutions.

In our example, when calculating the discriminant for the equation \(x^2 - x + 1 = 0\), we used \(a = 1\), \(b = -1\), and \(c = 1\). Plugging these into the formula gives:\[(-1)^2 - 4 \times 1 \times 1 = 1 - 4 = -3\]Since the discriminant is negative (-3), we conclude that the equation has no real solutions, but instead, two complex solutions exist.

Complex Solutions

When a quadratic equation's discriminant is negative, it cannot have real number solutions, instead, the solutions are complex. Complex numbers are numbers that include the imaginary unit \(i\), where \(i^2 = -1\).Complex solutions occur in pairs, as conjugates. A conjugate in this context means if one solution is \(a + bi\), the other will be \(a - bi\). This ensures that quadratics with real coefficients still distribute roots symmetrically around the real axis.For our problem \(x^2 - x + 1 = 0\), we calculate the solutions using the quadratic formula and find:\[x = \frac{1 \pm i\sqrt{3}}{2}\]These results, \(\frac{1 + i\sqrt{3}}{2}\) and \(\frac{1 - i\sqrt{3}}{2}\), are the conjugate complex solutions. They include the imaginary part, \(\pm i\sqrt{3}\), which arises directly from the negative discriminant (-3) within the square root.

Quadratic Formula

The quadratic formula provides a reliable way to solve any quadratic equation. It allows us to find solutions, whether they are real or complex. The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Understanding its components:

• \(-b\): This is the opposite of the coefficient \(b\) from the equation.
• \(\pm\): This symbol shows there will typically be two solutions, forming the roots.
• \(\sqrt{b^2 - 4ac}\): This part contains the discriminant and determines the nature of the roots.
• \(2a\): This part is the denominator that balances the equation based on the leading coefficient \(a\).

In the given equation \(x^2 - x + 1 = 0\), substituting values \(a = 1\), \(b = -1\), and \(c = 1\) into the formula gives us:\[x = \frac{-(-1) \pm \sqrt{-3}}{2 \times 1} = \frac{1 \pm i\sqrt{3}}{2}\]Thus, the quadratic formula reveals the complex nature of the solutions. It skillfully navigates us through the process regardless of whether the equation presents real or complex roots.