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Chapter 1: Problem 26

Solve by using the square root property. \(6 v^{2}-30=0\)

Short Answer

Expert verified
The solutions are \(v = \pm\sqrt{5}\).

Step by step solution

01

Move the constant to the other side

To isolate the variable term, add 30 to both sides of the equation: \[6v^2 - 30 + 30 = 0 + 30\] Which simplifies to: \[6v^2 = 30\]
02

Isolate the squared term

Divide both sides by 6 to isolate the squared term: \[\frac{6v^2}{6} = \frac{30}{6}\] This simplifies to: \[v^2 = 5\]
03

Apply the square root property

Take the square root of both sides of the equation to solve for v: \[v = \pm\sqrt{5}\]

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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 technique used to solve quadratic equations. When you have an equation set up in the form \[x^2 = k\], you can solve for x by taking the square root of both sides. This results in two possible solutions because both the positive and negative square roots of k are valid.

In mathematical terms, if you have \[v^2 = 5\], then \[v = \pm \sqrt{5}\].

This method is particularly useful for equations that can be easily isolated to have a squared term equal a constant. It simplifies the process and avoids the need for more complicated methods like factoring or using the quadratic formula.
Isolating Variables
Isolating variables means manipulating the equation so that the variable you are solving for is by itself on one side of the equation.

In the quadratic equation \ \[6v^2 - 30 = 0\ \], our goal is to isolate \[v^2\ \]. Here's how we do it step-by-step:
  • First, add 30 to both sides: \ \[6v^2 - 30 + 30 = 0 + 30\ \]
  • This simplifies to \ \[6v^2 = 30\ \]
  • Next, divide both sides by 6 to get \[v^2\ \] by itself: \ \[\frac{6v^2}{6} = \frac{30}{6}\ \]
  • That results in \[v^2 = 5\ \]

By eliminating any constants and coefficients from around the variable, you make it easier to apply further mathematical techniques, like the square root property or other methods for solving equations.
Simplifying Equations
Simplifying equations involves performing operations to make an equation more manageable. It's like cleaning up the equation to get it into its simplest form.

For example, in the original equation \[6v^2 - 30 = 0\], the simplification steps are:
  • Add 30 to both sides to get \[6v^2 = 30\]
  • Divide both sides by 6 resulting in \[v^2 = 5\]

Each of these steps removes unnecessary complexity and prepares the equation for final solving. This process is crucial because it often turns a difficult problem into something much easier to handle.

As you can see, simplifying an equation isn't just about making the math easier; it's about making it possible to solve with fewer steps.

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Most popular questions from this chapter

Solve the equations. \(6=7+|9 z-3|\)

Martha is in a boat in the ocean \(48 \mathrm{mi}\) from point \(A\), the closest point along a straight shoreline. She needs to dock the boat at a marina \(x\) miles farther up the coast, and then drive along the coast to point \(B, 96 \mathrm{mi}\) from point \(A\). Her boat travels \(20 \mathrm{mph}\), and she drives \(60 \mathrm{mph}\). If the total trip took \(4 \mathrm{hr}\), determine the distance \(x\) along the shoreline.

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