Problem 49
Question
Find and correct the error at the right. $$ \begin{array}{r} {x^{2}+36=0} \\ {x^{2} =-36} \\ {x=\pm 6} \end{array} $$
Step-by-Step Solution
Verified Answer
The correct solution should have been \(x = ±6i\) instead of \(x = ±6\). The mistake was made when taking the square root of -36.
1Step 1: Explore the initial equation
Start with the initial equation which is \(x^{2} + 36 = 0\). To isolate x squared, subtract 36 from both sides of the equation to get \(x^{2} = -36\). This bit of the calculation is done right.
2Step 2: Identify the erroneous step
Acknowledge that the error lies in the final square root step. From the equation \(x^{2} = -36\), it is incorrect to state that \(x = ±6\), as it implies taking the square root of -36 which does not result in real numbers.
3Step 3: Correcting the erroneous step
The correct step should involve understanding the square root of -36 falls within the realm of complex mathematics, resulting in imaginary numbers. To find \(x\), it is necessary to take the square root of both sides. The square root of -36 is ±6i, where 'i' is the imaginary unit with the property \(i^{2} = -1\). So the correct \(x\) should be \(x = ±6i\).
Key Concepts
Complex NumbersImaginary NumbersSquare Root of Negative Numbers
Complex Numbers
Complex numbers are essential in mathematics because they allow us to solve equations that have no real number solutions. A complex number comprises a real part and an imaginary part, and it takes the general form of \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit.
- The real part \( a \) is simply a real number that you are already familiar with.
- The imaginary part \( bi \) incorporates the imaginary unit \( i \).
Imaginary Numbers
Imaginary numbers extend the concept of complex numbers by focusing on the imaginary unit \( i \), defined such that \( i^2 = -1 \). These numbers are not imaginary in the sense of not existing but rather conceptual tools that enhance the framework of algebraic solutions. They allow us to address negative roots and radically shift equations in fields like electrical engineering and control systems.
- Imaginary numbers are particularly crucial when the discriminant of a quadratic equation is negative.
- They essentially rotate the number line into a plane, providing a complete set of solutions.
Square Root of Negative Numbers
Taking the square root of negative numbers once seemed like an unsolvable riddle in mathematics. This is where imaginary numbers come into play, as they offer a path forward when dealing with negatives under the radical. The square root of a negative number, \( -n \), is expressed as \( \sqrt{-n} = \sqrt{n} \cdot i \).
- In our context, the square root of \(-36\) becomes \( \pm 6i \) because \( \sqrt{36} = 6 \), and is multiplied by \( i \).
- This transforms the problem into working with complex solutions rather than calling it unsolvable.
Other exercises in this chapter
Problem 49
Simplify the expression. $$ \sqrt{\frac{1}{2}} $$
View solution Problem 49
Determine whether the number is a perfect square. $$ -4 $$
View solution Problem 50
Graph the exponential function. (Lesson 8.3) $$ y=\left(\frac{1}{3}\right)^{x} $$
View solution Problem 50
In Exercises \(49-51\), sketch the graphs of the three functions in the same coordinate plane. Then describe how the three parabolas are similar to each other a
View solution