Using Square Roots in Applications

Learning Outcomes

Solve application problems that include the use of square roots

Use Square Roots in Applications

As you progress through your college courses, you’ll encounter several applications of square roots. Once again, if we use our strategy for applications, it will give us a plan for finding the answer!

Use a strategy for applications with square roots.

Identify what you are asked to find.
Write a phrase that gives the information to find it.
Translate the phrase to an expression.
Simplify the expression.
Write a complete sentence that answers the question.

Square Roots and Area

We have solved applications with area before. If we were given the length of the sides of a square, we could find its area by squaring the length of its sides. Now we can find the length of the sides of a square if we are given the area, by finding the square root of the area.
If the area of the square is $A$ square units, the length of a side is $\sqrt{A}$ units. See the table below.

Area (square units)	Length of side (units)
$9$	$\sqrt{9}=3$
$144$	$\sqrt{144}=12$
$A$	$\sqrt{A}$

example

Mike and Lychelle want to make a square patio. They have enough concrete for an area of $200$ square feet. To the nearest tenth of a foot, how long can a side of their square patio be?

Solution
We know the area of the square is $200$ square feet and want to find the length of the side. If the area of the square is $A$ square units, the length of a side is $\sqrt{A}$ units.

What are you asked to find?	The length of each side of a square patio
Write a phrase.	The length of a side
Translate to an expression.	$\sqrt{A}$
Evaluate $\sqrt{A}$ when $A=200$ .	$\sqrt{200}$
Use your calculator.	$14.142135..$ .
Round to one decimal place.	$\text{14.1 feet}$
Write a sentence.	Each side of the patio should be $14.1$ feet.

try it

Square Roots and Gravity

Another application of square roots involves gravity. On Earth, if an object is dropped from a height of $h$ feet, the time in seconds it will take to reach the ground is found by evaluating the expression ${\Large\frac{\sqrt{h}}{4}}$ . For example, if an object is dropped from a height of $64$ feet, we can find the time it takes to reach the ground by evaluating ${\Large\frac{\sqrt{64}}{4}}$ .

	${\Large\frac{\sqrt{64}}{4}}$
Take the square root of $64$ .	${\Large\frac{8}{4}}$
Simplify the fraction.	$2$

It would take $2$ seconds for an object dropped from a height of $64$ feet to reach the ground.

example

Christy dropped her sunglasses from a bridge $400$ feet above a river. How many seconds does it take for the sunglasses to reach the river?

Show Solution

Solution

What are you asked to find?	The number of seconds it takes for the sunglasses to reach the river
Write a phrase.	The time it will take to reach the river
Translate to an expression.	${\Large\frac{\sqrt{h}}{4}}$
Evaluate $\Large\frac{\sqrt{h}}{4}$ when $h=400$ .	${\Large\frac{\sqrt{400}}{4}}$
Find the square root of $400$ .	${\Large\frac{20}{4}}$
Simplify.	$5$
Write a sentence.	It will take $5$ seconds for the sunglasses to reach the river.

try it

Square Roots and Accident Investigations

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes. According to some formulas, if the length of the skid marks is $d$ feet, then the speed of the car can be found by evaluating $\sqrt{24d}$ .

example

After a car accident, the skid marks for one car measured $190$ feet. To the nearest tenth, what was the speed of the car (in mph) before the brakes were applied?

Show Solution

Solution

What are you asked to find?	The speed of the car before the brakes were applied
Write a phrase.	The speed of the car
Translate to an expression.	$\sqrt{24d}$
Evaluate $\sqrt{24d}$ when $d=190$ .	$\sqrt{24\cdot 190}$
Multiply.	$\sqrt{4,560}$
Use your calculator.	$67.527772..$ .
Round to tenths.	$67.5$
Write a sentence.	The speed of the car was approximately $67.5$ miles per hour.

Module 6: Applications With Decimals