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Chapter 1: Problem 37

Solve the equation both algebraically and graphically. $$x^{2}+9=0$$

Short Answer

Expert verified
The solutions are complex: \(x = 3i\) and \(x = -3i\).

Step by step solution

01

Analyze the Equation Type

First, we observe that the equation is a quadratic equation in the form of \(x^2 + 9 = 0\). This equation does not have a linear term (the term with \(x\)) or a constant term other than 9.
02

Rearrange the Equation

To solve algebraically, move 9 to the right side of the equation: \(x^2 = -9\).
03

Assess Real Solutions

Notice that \(x^2 = -9\) suggests we must take the square root of a negative number, which indicates there are no real solutions. Instead, we proceed with complex solutions.
04

Solve Algebraically for Complex Solutions

Find the square root of both sides: \(x = \sqrt{-9}\). Using the properties of square roots, particularly dealing with negative numbers, this becomes \(x = \pm 3i\), where \(i\) is the imaginary unit (\(i^2 = -1\)).
05

Graphical Representation of the Equation

To solve graphically, recognize that the function \(f(x) = x^2 + 9\) is always above the x-axis, starting from \(y = 9\) when \(x = 0\). Since it never crosses the x-axis, there are no real x-intercepts, confirming no real solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
In mathematics, a quadratic equation is a polynomial equation of degree two. It generally has the standard form:
  • \[ ax^2 + bx + c = 0 \]
Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). They are known as the coefficients. Quadratic equations can be solved using various methods:
  • Factoring
  • Completing the square
  • Using the quadratic formula
  • Graphical methods
In the exercise provided, the quadratic equation \( x^2 + 9 = 0 \) lacks a linear term (the one with coefficient \( b \)) and hones in on identifying a special case due to its constant nature. With no real roots, such equations open the door to understanding complex numbers, which expand solutions beyond the real number line.
Imaginary Unit
The imaginary unit, represented as \( i \), is a fundamental concept in complex numbers. It is defined by the equation:
  • \( i^2 = -1 \)
This definition allows us to calculate the square roots of negative numbers, a task impossible in the realm of real numbers. In our exercise, to solve \( x^2 = -9 \), we leverage \( i \) to find the solutions. Here's how it works:
  • The expression \( \sqrt{-9} \) transforms by recognizing \( \sqrt{-1} = i \), leading to \( \sqrt{-9} = \sqrt{9} \cdot \sqrt{-1} = 3i \).
  • This gives the solutions \( x = 3i \) and \( x = -3i \).
The introduction of \( i \) offers an elegant solution to problems traditionally defined unsolvable under real numbers, enriching the mathematical landscape.
Complex Solutions
Complex solutions arise when quadratic equations like \( x^2 + 9 = 0 \) yield no real roots due to the presence of negative numbers under a square root. Complex numbers consist of a real part and an imaginary part, typically expressed in the form:
  • \( a + bi \), where \( a \) is the real component and \( bi \) is the imaginary component.
For our specific quadratic equation:
  • The solutions are \( x = \pm 3i \), pure imaginary numbers, since the real part is zero.
Understanding complex solutions involves recognizing:
  • These numbers lie off the real number line.
  • The graphical representation confirms this, as \( f(x) = x^2 + 9 \) never meets the x-axis, reinforcing the absence of real solutions.
  • Complex solutions allow us to employ a broader mathematical toolkit, solving more intricate problems by moving beyond conventional boundaries.
Harnessing the power of complex numbers empowers us to solve and graphically represent solutions far beyond what real numbers alone could handle.

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Most popular questions from this chapter

Factor the expression completely. $$4 x^{2}+4 x y+y^{2}$$

Factor the expression completely. $$30 x^{3}+15 x^{4}$$

Factor the expression completely. $$12 x^{3}+18 x$$

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Verify these formulas by expanding and simplifying the right-hand side. $$ \begin{array}{l} A^{2}-1=(A-1)(A+1) \\ A^{3}-1=(A-1)\left(A^{2}+A+1\right) \\ A^{4}-1=(A-1)\left(A^{3}+A^{2}+A+1\right) \end{array} $$ On the basis of the pattern displayed in this list, how do you think \(A^{5}-1\) would factor? Verify your conjecture. Now generalize the pattern you have observed to obtain a factoring formula for \(A^{n}-1,\) where \(n\) is a positive integer.
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