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Chapter 1: Problem 79

Find the real solutions, if any, of each equation. Use any method. $$ x^{2}-5=0 $$

Short Answer

Expert verified
The solutions are \(x = \sqrt{5}\) and \(x = -\sqrt{5}\).

Step by step solution

01

Add 5 to both sides

To isolate the term with the variable, add 5 to both sides of the equation: \[ x^2 - 5 + 5 = 0 + 5 \]which simplifies to: \[ x^2 = 5 \]
02

Take the square root of both sides

To solve for \(x\), take the square root of both sides of the equation: \[ x = \pm \sqrt{5} \]This gives us the two possible solutions.

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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 polynomials that can be written in the form \( ax^2 + bx + c = 0 \). Here, 'a', 'b', and 'c' are constants and 'x' is the variable we want to solve for. In our example, the equation is \( x^2 - 5 = 0 \), which is already in the standard quadratic form with \( a = 1 \), \( b = 0 \), and \( c = -5 \). There are several methods to solve quadratic equations including factoring, completing the square, using the quadratic formula, and taking square roots. For simplicity, we will use the square root method here.
Square Root Method
The square root method is an effective way to solve quadratic equations when they are in a particular form. Specifically, this method works well if the equation can be rewritten as \( x^2 = k \), where 'k' is a constant.

In our example, the equation \( x^2 - 5 = 0 \) can be rearranged to \( x^2 = 5 \). Now, we need to isolate 'x'. The next step involves taking the square root of both sides of the equation.

When we take the square root of both sides, we must remember that the square root function returns both positive and negative roots. Therefore, the solutions to our equation become: \( x = \pm \sqrt{5} \).

This results in two possible solutions: \sqrt{5} and -\sqrt{5}.
Real Solutions
When solving quadratic equations, the 'real solutions' refer to solutions that are real numbers. These are different from imaginary or complex numbers. In this example, after applying the square root method, we arrived at \sqrt{5} and -\sqrt{5}.

Both of these numbers are real because they can be found on the number line. Real solutions are significant because they represent the actual values 'x' can take in real-world problems.

To summarize, the equation \( x^2 - 5 = 0 \) has two real solutions: \sqrt{5} and -\sqrt{5}. Being thorough with identifying and validating that these solutions are real helps ensure our problem-solving is accurate and meaningful in practical scenarios.

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Most popular questions from this chapter

Electrical Circuits The impedance \(Z,\) in ohms, of a circuit element is defined as the ratio of the phasor voltage \(V,\) in volts, across the element to the phasor current \(I\), in amperes, through the element. That is, \(Z=\frac{V}{I} .\) If the voltage across a circuit element is \(18+i\) volts and the current through the element is \(3-4 i\) amperes, determine the impedance.

Solve each inequality. Express your answer using set notation or interval notation. Graph the solution set. $$ |3 x| \geq 0 $$

Write each inequality using interval notation, and graph each inequality on the real mumber line. \(x<-4\)

Find the real solutions, if any, of each equation. Use any method. $$ 6 x^{2}+7 x-20=0 $$

True or False Radical equations sometimes have extraneous solutions.
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