Problem 5 - Basic Technical Mathematics with Calculus

Question

Find the roots of the given equations by inspection. $$\left(x^{2}+6 x+9\right)\left(x^{2}+4\right)=0$$

Step-by-Step Solution

Answer

The roots are \( x = -3, x = 2i, \) and \( x = -2i \).

Step 1: Understand the Equation

The given equation is\( \left(x^{2}+6x+9\right)\left(x^{2}+4\right)=0 \). It is a product of two separate equations: \( x^2 + 6x + 9 \) and \( x^2 + 4 \). The equation will be zero if either factor is zero.

Step 2: Solve for Roots of the First Factor

The first factor is \( x^2 + 6x + 9 \). This can be rewritten as \((x + 3)^2 = 0\). This means \( x + 3 = 0 \), giving the root \( x = -3 \).

Step 3: Solve for Roots of the Second Factor

The second factor is \( x^2 + 4 \). This equation can be rewritten as \( x^2 = -4 \). The solutions are complex numbers: \( x = \pm 2i \). These are complex roots because the square root of a negative number involves 'i', the imaginary unit.

Step 4: Conclusion of Roots

Combine the results: The roots of the entire equation are \( x = -3 \) from the first factor and \( x = 2i \) and \( x = -2i \) from the second factor.

Key Concepts

Quadratic Equation

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants and \( a eq 0 \). Quadratic equations always have two solutions, which could be real or complex numbers.
Understanding how to solve quadratic equations is fundamental in algebra. There are different methods to find the roots of a quadratic equation, including taking advantage of the quadratic formula, factoring, or simply inspecting simpler equations.

• Factoring: This is the easiest method if the quadratic can be factored easily, as was the case with \( x^2 + 6x + 9 \) since it can be rewritten as \((x + 3)^2 = 0\).
• Quadratic Formula: The solutions for a general quadratic equation \( ax^2 + bx + c = 0 \) are given by the formula: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]
• Completing the Square: Another method, where you complete the square on the quadratic expression to make it a perfect square trinomial.

Quadratic equations, therefore, can yield real solutions, like \( x = -3 \), or complex solutions, which we'll explore next.

Complex Numbers

When we have a quadratic equation with no real roots, such as \( x^2 + 4 = 0 \), it results in complex roots. Complex numbers are expressed in the form \( a + bi \), where \( a \) is the real part, \( b \) is the imaginary part, and \( i \) is the imaginary unit with the property that \( i^2 = -1 \).
In our equation, to find roots of \( x^2 = -4 \), you are essentially looking at taking the square root of a negative number:

• This becomes \( x = \pm \sqrt{-4} \).
• Rewriting it, we have \( x = \pm 2i \). This is because \( \sqrt{-4} = \sqrt{4} \times \sqrt{-1} = 2 \times i \).

Complex numbers are crucial in mathematics since they extend our ability to solve equations that do not have real solutions. They appear naturally in many areas of science and engineering, particularly in signal processing and control theory.

Factoring Polynomials

Factoring is the process of breaking down an equation or expression into simpler multiples called factors. It is particularly useful in solving polynomial equations.
For example, given \( x^2 + 6x + 9 \), you can factor it as \((x+3)^2\). This is essentially reversing the process of expansion.
Some tips for effective factoring are:

• Common Factor: Check if all terms have a greatest common factor (GCF) first.
• Trinomials: Check for patterns of perfect square trinomials or differences of squares.
• Grouping: For higher degree polynomials, sometimes grouping terms can aid in factoring.

In practice, applying factoring makes solving equations straightforward, by reducing the levels of complexity one step at a time. Factoring not only helps in finding solutions but also in understanding the structure of mathematical expressions, thus aiding in simplifying complex algebraic equations.