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Chapter 9: Problem 25

\(p^{2}=147\)

Short Answer

Expert verified
\( p = \pm 7 \sqrt{3} \)

Step by step solution

01

- Understand the Equation

The given equation is a simple quadratic equation in terms of the variable `p`: \( p^2 = 147 \). The goal is to solve for `p`.
02

- Isolate the Variable

To isolate `p`, take the square root of both sides of the equation. Remember that taking the square root of both sides introduces two solutions, one positive and one negative. \[ p = \pm \sqrt{147} \]
03

- Simplify the Square Root

Simplify \( \sqrt{147} \). Notice that 147 can be factored into 49 and 3, where 49 is a perfect square:\[ 147 = 49 \times 3 \]Therefore, \( \sqrt{147} = \sqrt{49 \times 3} = \sqrt{49} \times \sqrt{3} = 7 \sqrt{3} \)
04

- Write the Final Solutions

Hence, the solutions for `p` are:\[ p = \pm 7 \sqrt{3} \]

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Key Concepts

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

Square Root Property
To solve quadratic equations like the one in the exercise (\(p^2 = 147\)), we often use the square root property. This property allows us to isolate the variable and solve for it. The square root property states that \(x^2 = a\) can be solved by taking the square root of both sides, giving \(x = \pm \sqrt{a}\).

Here are the steps:
  • Identify the equation in the form \(x^2 = a\)
  • Take the square root of both sides, remembering to include both the positive and negative roots
For instance, for \(p^2 = 147\):
  • Take square root of both sides:
    \(p = \pm \sqrt{147}\)
This gives us the initial step towards solving the quadratic equation using the square root property. This property is not just for simple equations but also a stepping stone in more complicated algebra and calculus problems.
Factoring
Factoring is another key concept for solving quadratic equations, but in this particular exercise, it's used indirectly while simplifying the square root.

Factoring involves breaking down a number or an expression into its simplest building blocks (factors). For instance, the number 147 can be factored into 49 and 3 because \(147 = 49 \times 3\).

Using prime factorization:
  • 147 divided by 3 gives 49 (147 = 3 \times 49)
  • 49 is a square of 7 (49 = 7 \times 7)
Therefore, we can express:
\(147 = 3 \times 49\)n which can be written as:
\(147 = 3 \times 7^2\)Factoring helps simplify complex problems, and you frequently use it to make radical expressions easier to manage, as seen when we simplify \( \sqrt{147}\) in our solution.
Simplifying Radicals
Simplifying radicals is an important step in solving equations involving roots. A radical expression includes a square root, cube root, or any root. Simplifying radicals makes the expressions cleaner and easier to understand.

For example, let's simplify \( \sqrt{147}\) from our exercise:
  • We factored 147 into 49 and 3 because 49 is a perfect square and easier to handle inside a square root
  • This gives us: \( \sqrt{147} = \sqrt{49 \times 3}\)
  • Using the property of square roots (where \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\)), we can write:
    \( \sqrt{147} = \sqrt{49} \times \sqrt{3}\)
  • And since the square root of 49 is 7, it becomes:
    \( \sqrt{49} \times \sqrt{3} = 7 \sqrt{3}\)
Simplifying radicals involves recognizing these factorizations and using properties of roots to make expressions simpler. It helps solve equations and makes sure your final answer is in its simplest form, making it easier to interpret and use.

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Most popular questions from this chapter

The diameter at basal height of a tree is the diameter about \(3 \mathrm{ft}\) above the ground. The approximate shape of the trunk of a tree above basal height is a cone. The formula for the volume of a cone is \(V=\frac{\pi r^{2} h}{3}\), where \(r\) is the radius and \(h\) is the height. Measured from its basal height, a tree is \(50 \mathrm{ft}\) tall. The diameter of its trunk is \(2.5 \mathrm{ft}\). Find the approximate volume of lumber in cubic feet in this trunk. ( \(\pi \approx 3.14\).) Round to the nearest whole number. (Source: G. John Smith; www.math.bcit.ca, 1997)

\(x^{2}-3 x+3=0\)

\(x^{2}=-100\)

\(z^{2}+3 z+14=0\)

Explain why the square root of a negative number cannot be a real number.
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