Chapter 5.5, Problem 5CYU

To determine

To find: the distance that the projected laser light show had an area of 100 square meter.

Answer to Problem 5CYU

The distance from the laser to the screen in feet is $\text{25feet}$ .

Explanation of Solution

Given:

  $\text{A=0}{\text{.16d}}^2$ .

Where $\text{d}$ is distance from the laser to the screen.

Concept used:

The square root property is one method that can be used to solve the quadratic equations, this method is generally used on equations that have form ${\text{ax}}^{\text{2}}\text{+bx+c}\text{or}{\left(\text{ax+b}\right)}^{\text{2}}\text{=c}$ ,or an equation that can be expressed in either or those forms.

All need to do is take a square root of both sides and solved for the variable.

Calculation:

Use the square root property in order to solve the equation:

  $\text{0}{\text{.16d}}^2=100$ .

  ${\left(\text{0}\text{.4d}\right)}^2={10}^2$ .

Using square root property:

  $\sqrt{{\left(\text{0}\text{.4d}\right)}^2}=\pm \sqrt{{10}^2}$ .

  $\text{0}\text{.4d}=\pm 10$ .

Dividing both side by $0.4$ :

  $\text{d=}\frac{\text{10}}{\text{0}\text{.4}}$ or $\text{d=-}\frac{\text{10}}{\text{0}\text{.4}}$ .

  $\text{d=25}$ and $\text{d=-25}$ .

Hence the distance from the laser to the screen in feet is $\text{25feet}$ .