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Chapter 11: Problem 80

Solve each equation. (All solutions are nonreal complex numbers.) $$ (x-5)^{2}=-36 $$

Short Answer

Expert verified
The solutions are \(x = 5 + 6i\) and \(x = 5 - 6i\).

Step by step solution

01

Isolate the squared term

To begin solving \((x-5)^2 = -36\), first note that the squared term \((x-5)^2\) is already isolated on one side of the equation.
02

Take the square root of both sides

Take the square root of both sides of the equation to eliminate the exponent. Remember to include the \( \pm \) symbol when taking the square root of both sides: \(\text{sqrt}( (x-5)^2 ) = \pm \text{sqrt}( -36 )\). This simplifies to \(x - 5 = \pm 6i\) since \(\text{sqrt}(-36) = 6i\)
03

Solve for x

Now, solve for \(x\) by isolating it on one side of the equation: \(x - 5 = 6i\) and \(x - 5 = -6i\). This yields two solutions: \(x = 5 + 6i\) and \(x = 5 - 6i\).

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Key Concepts

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

Solving Quadratic Equations
Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). To solve these equations, we look for values of \(x\) that satisfy the equation.
Sometimes, the quadratic equation may be set up in a way similar to the given problem, \((x-5)^{2}=-36.\)
This form can often be easier to work with. The main steps involve isolating the term with the variable, taking roots, and then solving for the variable. Let's see how this works in detail:
  • First, we isolate the quadratic term.
  • Then, we take the square root of both sides to remove the square.
  • Lastly, we solve for the variable to find all possible solutions.
In this specific equation, we end up with two complex solutions, demonstrating another important concept: complex numbers.
Complex Numbers
Complex numbers are an extension of the real numbers and are particularly useful when dealing with square roots of negative numbers. A complex number is typically written as \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property \(i^{2} = -1.\)
In this work, we started with the equation \((x-5)^{2}=-36.\)
Taking the square root of a negative number leads us to an imaginary number. The square roots of \(-36\) are \(\text{6i and -6i}\). This leads us to the two possible values for \(x\): \(5 + 6i\) and \(5 - 6i.\)
These solutions illustrate how complex numbers can represent values that can't be expressed on the real number line alone.
Square Roots of Negative Numbers
In basic algebra, the square of any real number is always non-negative. However, when solving equations like our given quadratic, we encounter the need to take the square root of a negative number.
When we try to take the square root of \(-36\), we recognize that no real number squared gives \(-36\). This is where imaginary numbers come into play:
  • The square root of \(-1\) is defined to be \(i\), an imaginary unit.
  • Then, the square root of \(-36\) is calculated as \(6i\), because \(\text{6} \times \text{i} = 6i\) and \((6i)^2 = 36 \times -1 = -36.\)
With this tool, you can find solutions to equations that wouldn't have solutions in the realm of real numbers.

Understanding these square roots and their properties is crucial for dealing with complex number equations.

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Most popular questions from this chapter

The following exercises are not grouped by type. Solve each equation. (Exercises 83 and 84 require knowledge of complex numbers.) \(2 x^{4}+x^{2}-3=0\)

Solve each equation by completing the square. Give (a) exact solutions and (b) solutions rounded to the nearest thousandth. \(3 r^{2}-2=6 r+3\)

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ x^{2}+18=10 x $$

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ 4 r^{2}-4 r-19=0 $$

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ (2 x-1)(8 x-4)=-1 $$
See all solutions

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