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Chapter 1: Problem 19

Solve each equation in Exercises \(15-34\) by the square root property. $$ 2 x^{2}-5=-55 $$

Short Answer

Expert verified
The solutions to the equation are \(x = 5i\) and \(x = -5i\).

Step by step solution

01

Rearranging the equation

First, isolate \(x^2\) on one side of the equation. Given \(2x^2 - 5 = -55\), add \(5\) to both sides to get \(2x^2 = -50\). Then divide both sides by \(2\) to get \(x^2 = -25\).
02

Applying the Square Root Property

Once the equation is in the form \(x^2 = k\), apply the square root property. The square root of a number encompasses both positive and negative roots. Therefore, \(x = \sqrt{-25}\) or \(x = -\sqrt{-25}\).
03

Simplifying the Results

We need to simplify the square root. The square root of \(-25\) is an imaginary number. We represent it as \(5i\), where \(i = \sqrt{-1}\). Therefore, the solutions to the equation are \(x = 5i\) and \(x = -5i\).

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Key Concepts

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

Square Root Property
The square root property is a crucial concept when it comes to solving quadratic equations in the form \(x^2 = k\). This property states that if \(x^2 = k\), then \(x = \pm \sqrt{k}\). Simply put, this means when we solve for \(x\), we must consider both the positive and negative square roots of \(k\).

To effectively use this property:
  • First isolate \(x^2\) on one side of the equation. This means the equation should appear as \(x^2 = k\).
  • Once \(x^2\) is isolated, apply the square root to both sides.
  • Remember to include both \(\sqrt{k}\) and \(-\sqrt{k}\) as potential solutions.

In our example, starting with \(2x^2 - 5 = -55\), we re-arranged it to \(x^2 = -25\) and then applied the square root property to find possible values for \(x\). This is when we encounter the need to work with imaginary numbers.
Imaginary Numbers
Imaginary numbers come into play when you have the square root of a negative number. In mathematics, it's not possible to have the square root of a negative number result in a real number. That's where imaginary numbers, represented by \(i\), are useful. The imaginary unit \(i\) is defined as \(i = \sqrt{-1}\).

Here's how imaginary numbers are used:
  • To rewrite the square root of a negative number, factor out \(-1\).
  • For instance, \(\sqrt{-25}\) becomes \(\sqrt{-1} \cdot \sqrt{25} = i \cdot 5 = 5i\).

In the step-by-step exercise, \(\sqrt{-25}\) resulted in the imaginary number \(5i\). Imaginary numbers like \(5i\) and \(-5i\) are now considered distinct solutions of the equation.
Complex Solutions
In the context of quadratic equations, complex solutions often arise when the discriminant (the expression under the square root in the quadratic formula) is negative, indicating that there are no real roots. The solutions to these equations are instead expressed as complex numbers.

A complex number takes the form \(a + bi\), where:
  • \(a\) is the real part.
  • \(b\) is the imaginary part.

In our example, the equation \(x^2 = -25\) leads us directly to complex solutions. By applying the square root property, we found the solutions \(x = 5i\) and \(x = -5i\), which fall under the umbrella of complex numbers. These complex solutions are crucial in understanding behaviors in applied fields like engineering and physics.
Rearranging Equations
Rearranging equations is often the first step in solving them, especially in preparations for applying methods like the square root property. This process ensures that the equation is in a form suitable for further analysis.

To effectively rearrange an equation:
  • Identify the term with \(x^2\) and work to isolate it on one side of the equation.
  • Use addition or subtraction to move all other terms to the opposite side.
  • Utilize multiplication or division to further simplify the equation, if necessary.

For instance, in our exercise, we started with the equation \(2x^2 - 5 = -55\). By adding \(5\) to both sides, we eliminated the constant term, resulting in \(2x^2 = -50\). Dividing both sides by \(2\) gave us \(x^2 = -25\), setting up a clearer path to apply the square root property. This strategic rearrangement is essential to accurately solve quadratic equations.

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Most popular questions from this chapter

In Exercises \(127-130,\) solve each equation by the method of your choice. $$ \frac{x-1}{x-2}+\frac{x}{x-3}=\frac{1}{x^{2}-5 x+6} $$

In a film, the actor Charles Coburn plays an elderly "uncle" character criticized for marrying a woman when he is 3 times her age. He wittily replies, "Ah, but in 20 years time I shall only be twice her age." How old is the "uncle" and the woman?

A tennis club offers two payment options. Members can pay a monthly fee of \(\$ 30\) plus \(\$ 5\) per hour for court rental time. The second option has no monthly fee, but court time costs \(\$ 7.50\) per hour. a. Write a mathematical model representing total monthly costs for each option for \(x\) hours of court rental time. b. Use a graphing utility to graph the two models in a \([0,15,1]\) by \([0,120,20]\) viewing rectangle. c. Use your utility's trace or intersection feature to determine where the two graphs intersect. Describe what the coordinates of this intersection point represent in practical terms. d. Verify part (c) using an algebraic approach by setting the two models equal to one another and determining how many hours one has to rent the court so that the two plans result in identical monthly costs.

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