Problem 70 - Precalculus Mathematics for Calculus

Question

Find all solutions of the equation and express them in the form \(a+b i\) $$4 x^{2}-16 x+19=0$$

Step-by-Step Solution

Answer

The solutions are \(x = 2 + \frac{\sqrt{3}}{2}i\) and \(x = 2 - \frac{\sqrt{3}}{2}i\).

Step 1: Identify a, b, and c in the quadratic equation

The given quadratic equation is in the standard form \(ax^2 + bx + c = 0\). Compare the given equation \(4x^2 - 16x + 19 = 0\) with the standard form to identify \(a = 4\), \(b = -16\), and \(c = 19\).

Step 2: Calculate the discriminant

The discriminant (\(D\)) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(D = b^2 - 4ac\). Substitute the values of \(a\), \(b\), and \(c\) into the formula: \(D = (-16)^2 - 4 \cdot 4 \cdot 19\). Calculate this to get \(D = 256 - 304 = -48\).

Step 3: Determine the nature of the roots

Since the discriminant \(D = -48\) is less than zero, the quadratic equation has two complex conjugate solutions. We will use the quadratic formula to find these complex solutions.

Step 4: Use the quadratic formula

The quadratic formula to find the roots is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Plug in the values of \(a = 4\), \(b = -16\), and \(D = -48\). This gives us: \[x = \frac{-(-16) \pm \sqrt{-48}}{2 \cdot 4}\]. Simplify the equation to get \[x = \frac{16 \pm \sqrt{-48}}{8}\].

Step 5: Simplify the expression

Complete the calculation: \(x = \frac{16 \pm i\sqrt{48}}{8}\). Simplify \(\sqrt{48}\) as \(\sqrt{48} = 4\sqrt{3}\), then the expression becomes \[x = \frac{16 \pm 4i\sqrt{3}}{8}\]. Separate into two terms: \[x = 2 \pm \frac{i\sqrt{3}}{2}\].

Step 6: Express solutions in the form a+bi

The roots can now be clearly expressed in the form \(a + bi\). The solutions are: \(x_1 = 2 + \frac{\sqrt{3}}{2}i\) and \(x_2 = 2 - \frac{\sqrt{3}}{2}i\).

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental mathematical expression that can be written in the standard form as \(ax^2 + bx + c = 0\). It typically has three components: \(a\), \(b\), and \(c\), which are coefficients of the equation, with \(a\) not equal to zero.
Quadratic equations are called such because the highest power of the variable \(x\) is 2, making it a "quadratic" term.
Understanding a quadratic equation involves identifying the values of \(a\), \(b\), and \(c\) from the equation and interpreting what each part contributes to the shape and location of the curve represented in its graph.

• \(ax^2\): represents the parabola's opening direction and width; positive \(a\) means it opens upwards, negative \(a\) downwards.
• \(bx\): affects the axis of symmetry and the slope of the parabola.
• \(c\): represents the y-intercept, or where the parabola crosses the y-axis.

Quadratic equations may have real, repeated real, or complex solutions depending on the discriminant value.

Discriminant

The discriminant is a vital part of the quadratic equation's anatomy. It determines the nature of the roots without solving the equation entirely. The discriminant \(D\) is calculated using the formula: \(D = b^2 - 4ac\).
The value of \(D\) helps us understand the type of solutions we should expect:

• If \(D > 0\): The quadratic equation has two distinct real roots.
• If \(D = 0\): There is exactly one real root, also described as a repeated or double root.
• If \(D < 0\): The quadratic equation has no real roots, but rather two complex conjugate solutions.

By calculating the discriminant at the beginning of the problem-solving process, students can predict and prepare for dealing with potential complex number solutions.

Quadratic Formula

The quadratic formula is a universal method for finding the roots of any quadratic equation. Given as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), it offers a straightforward way to solve even the trickiest quadratic equations when factoring isn't feasible.
This formula works universally because it is derived from completing the square and is applicable regardless of the type of coefficient values (whether they are real, rational, etc.).

• Inside the formula, \(-b\) suggests the formula begins by taking the additive inverse of the linear coefficient \(b\).
• The \(\pm\) sign indicates the presence of two potential solutions, corresponding to adding or subtracting the square root of the discriminant.
• The denominator \(2a\) normalizes the calculated solutions over the double of the quadratic coefficient.

By using the quadratic formula, students can directly reach the solution of a quadratic equation, informed by the nature of the discriminant, whether real or involving complex numbers.