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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

Comparing Tablets The screen size of a tablet is determined by the length of the diagonal of the rectangular screen. The 12.9-inch iPad Pro \({ }^{\mathrm{TM}}\) comes in a 16: 9 format, which means that the ratio of the length to the width of the rectangular screen is \(16: 9 .\) What is the area of the iPad's screen? What is the area of a 12.3 -inch Microsoft Surface Pro \({ }^{\mathrm{TM}}\) if its screen is in a 3: 2 format? Which screen is larger? (Hint: If \(x\) is the length of a 4: 3 format screen, then \(\frac{3}{4} x\) is the width. \()\)

Find the real solutions, if any, of each equation. $$ \left(\frac{v}{v+1}\right)^{2}+\frac{2 v}{v+1}=8 $$

Challenge Problem If \(\frac{2}{3} \leq \frac{5-x}{3} \leq 3,\) find the largest possible value of \(2 x^{2}-3\)

The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If \(t_{1}\) is the time (measured in seconds) that it takes for the object to strike the water, then \(t_{1}\) will obey the equation \(s=16 t_{1}^{2}\), where \(s\) is the distance (measured in feet). It follows that \(t_{1}=\frac{\sqrt{s}}{4}\). Suppose that \(t_{2}\) is the time that it takes for the sound of the impact to reach your ears. Because sound waves are known to travel at a speed of approximately 1100 feet per second, the time \(t_{2}\) to travel the distance \(s\) will be \(t_{2}=\frac{s}{1100} .\) See the illustration. Now \(t_{1}+t_{2}\) is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have the equation $$ \text { Total time elapsed }=\frac{\sqrt{s}}{4}+\frac{s}{1100} $$ Find the distance to the water's surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds.

Challenge Problem Show that the real solutions of the equation \(a x^{2}+b x+c=0, a \neq 0,\) are the negatives of the real solutions of the equation \(a x^{2}-b x+c=0\). Assume that \(b^{2}-4 a c \geq 0\)
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