Problem 110 - Calculus for Biology and Medicine

Question

First determine whether the solutions of each quadratic equation are real or complex without solving the equation. Then solve the equation. \(4 x^{2}-x+1=0\)

Step-by-Step Solution

Answer

The quadratic equation has two complex conjugate solutions: \(x_1 = \frac{1 + i\sqrt{15}}{8}\) and \(x_2 = \frac{1 - i\sqrt{15}}{8}\).

Step 1: Finding the Discriminant

The discriminant of a quadratic equation of the form \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\). Here, \(a = 4\), \(b = -1\), \(c = 1\). Calculate the discriminant: \[ (-1)^2 - 4 \cdot 4 \cdot 1 = 1 - 16 = -15 \]

Step 2: Determining the Nature of the Solutions

Since the discriminant is negative, \(-15\), the solutions of the quadratic equation are complex (non-real). A negative discriminant indicates that the equation has two complex conjugate solutions.

Step 3: Solving the Quadratic Equation using the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Use the formula to find the solutions:\(x = \frac{-(-1) \pm \sqrt{-15}}{2 \cdot 4}\)Simplify the expression:\(x = \frac{1 \pm i\sqrt{15}}{8}\)

Step 4: Expressing the Solutions

The solutions can be written as two complex numbers:\(x_1 = \frac{1 + i\sqrt{15}}{8}\) and \(x_2 = \frac{1 - i\sqrt{15}}{8}\)

Step 5: Conclusion

The quadratic equation \(4x^2 - x + 1 = 0\) has two complex solutions: \(x_1 = \frac{1 + i\sqrt{15}}{8}\) and \(x_2 = \frac{1 - i\sqrt{15}}{8}\).

Key Concepts

Discriminant

In any quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant plays a crucial role in determining the nature of the roots without solving the equation. The discriminant is expressed as \(b^2 - 4ac\). It's a simple calculation with vital implications.

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it is zero, there is exactly one real root, which means the equation has a double root or repeated real solution.
• And importantly, if the discriminant is negative, like in our exercise with \(-15\), it indicates the equation has no real roots and instead will yield two complex conjugate solutions.

Being able to swiftly calculate and interpret the discriminant is helpful for quickly assessing whether solutions will be real or complex. The sign of the discriminant directs us to the next steps in learning about the solutions of the equation.

Complex Solutions

Complex solutions arise in quadratic equations when the discriminant is negative. You might wonder, what exactly are complex solutions and how do they relate to imaginary numbers?
Complex solutions typically have the form \(a + bi\) and \(a - bi\), where \(i\) is the imaginary unit defined by the property \(i^2 = -1\). These solutions occur in conjugate pairs because when we solve the quadratic equation using the quadratic formula, any square root of a negative number involves \(i\).
For example, from our quadratic equation \(4x^2 - x + 1 = 0\), the discriminant \(-15\) leads to solutions with imaginary parts, represented as \(\frac{1 \pm i\sqrt{15}}{8}\). Here, the real part is \(\frac{1}{8}\) and the imaginary part is \(\frac{\sqrt{15}}{8}\). Notice the two solutions are conjugates: \(\frac{1 + i\sqrt{15}}{8}\) and \(\frac{1 - i\sqrt{15}}{8}\). This conjugate pair property ensures that polynomial equations with real coefficients have complex roots in pairs.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of any quadratic equation \(ax^2 + bx + c = 0\). The formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

• It efficiently handles cases where the roots may be real or complex, determined by the discriminant \(b^2 - 4ac\).
• The expression within the square root, known as the discriminant, signals the nature of the solutions: positive for real, zero for a repeated real root, or negative for complex solutions.
• The formula is derived from the method of completing the square, providing a guaranteed route to solving quadratics when factoring isn't readily apparent or possible.

In our exercise, applying the quadratic formula to \(4x^2 - x + 1 = 0\) with a calculated discriminant of \(-15\) shows how straightforward it is to find the solutions when substituting into the formula: \(x = \frac{1 \pm i\sqrt{15}}{8}\). This step-by-step approach ensures accuracy in finding both real and complex roots.