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Chapter 0: Problem 23

Solve using the square root method. $$p^{2}-8=0$$

Short Answer

Expert verified
The solutions are \( p = 2\sqrt{2} \) and \( p = -2\sqrt{2} \).

Step by step solution

01

Isolate the Variable Term

The first step is to get the quadratic term \( p^2 \) by itself. To do this, add 8 to both sides of the equation: \[ p^2 - 8 + 8 = 0 + 8 \] This simplifies to: \[ p^2 = 8 \]
02

Apply the Square Root Method

Take the square root of both sides of the equation to solve for \( p \): \[ \sqrt{p^2} = \sqrt{8} \] Simplifying gives: \[ p = \pm \sqrt{8} \]
03

Simplify the Square Root

Simplify \( \sqrt{8} \): \[ \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2 \sqrt{2} \]Thus, \( p = \pm 2\sqrt{2} \).

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Key Concepts

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

Square Root Method
The square root method is a nifty way to solve quadratic equations that are in the form where the quadratic term is isolated on one side of the equation. What does this mean exactly?

Imagine you have a basic equation such as \( p^2 = a \). To find \( p \), you need to undo the squaring of \( p \). That's where the square root comes in handy.

By taking the square root of both sides, you can simplify to find the value of \( p \). Here's a tiny trick to remember:
  • When you take the square root of something like \( p^2 \), you find two potential solutions: \( p = \pm\sqrt{a} \).
  • The "卤" sign is crucial. It reminds you that both positive and negative roots are possible.
This makes the square root method a quick and easy tool for solving certain types of quadratic equations.
Isolating Variable Terms
Isolating variable terms is often your first move when tackling algebraic problems. Think of it as preparing the equation so you can directly use methods like the square root method. Here's how you do it:

The aim is to get the term with your variable鈥攊n our case, \( p^2 \)鈥攂y itself on one side of the equation.

To do this, you'll need to:
  • Rearrange or "move" constants or other numbers away from the variable part. For example, if you have \( p^2 - 8 = 0 \), you need to add 8 to both sides to isolate \( p^2 \) like so: \( p^2 = 8 \).
Why add 8, you ask? Because it does the opposite of subtracting 8, essentially neutralizing the effects on \( p^2 \). This leaves \( p^2 \) sitting alone, ready for the next step.
Simplifying Square Roots
When you encounter square roots like \( \sqrt{8} \), it's often helpful to simplify them further, especially when your final answer should be as neat as possible.

Here's a handy trick: Try to break down the number inside the square root sign into its prime factors. For instance:
  • \( \sqrt{8} \) can be split into \( \sqrt{4 \times 2} \).
  • Since \( \sqrt{4} \) is an easy-to-remember number (\( 2 \)), you simplify to \( 2\sqrt{2} \).
This process helps ensure your answers are in the simplest form possible and might even make other calculations down the line smoother. Plus, it strengthens your number sense鈥攎aking you more adept at noticing patterns and shortcuts in calculations.

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Most popular questions from this chapter

Solve the equation \(x^{1 / 2}=-4 x^{1 / 4}+21 .\) Plot both sides of the equation in the same viewing screen, \(y_{1}=x^{1 / 2}\) and \(y_{2}=-4 x^{1 / 4}+21 .\) Does the point(s) of intersection agree with your solution?

Determine whether the lines are parallel, perpendicular, or neither, and then graph both lines in the same viewing screen using a graphing utility to confirm your answer. $$\begin{aligned} &y_{1}=\frac{1}{2} x+5\\\ &y_{2}=2 x-3 \end{aligned}$$

Monthly Driving Costs. The monthly costs associated with driving a new Honda Accord are the monthly loan payment plus \(\$ 25\) every time you fill up with gasoline. If you fill up 5 times in a month, your total monthly cost is \(\$ 500 .\) How much is your loan payment?

Solve the equation \(\sqrt{x-3}=4-\sqrt{x+2} .\) Plot both sides of the equation in the same viewing screen, \(y_{1}=\sqrt{x-3}\) and \(y_{2}=4-\sqrt{x+2},\) and zoom in on the \(x\) -coordinate of the point of intersection. Does the graph agree with your solution?

Find an equation of a line that passes through the point \((-A, B-1)\) and is perpendicular to the line \(A x+B y=C\) Assume that \(A\) and \(B\) are both nonzero.
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