Beginning Algebra, 11th Edition

(Marvins-Underground-K-12) #1

Step 3 Write an equation.Use the Pythagorean theorem.


x^2 + 1 x+ 122 = 1 x+ 222


a^2 +b^2 =c^2


SECTION 6.6 Applications of Quadratic Equations 403


x

x+1

x+2

North

90 ° East

FIGURE 3

Be careful to
substitute properly.

Step 4 Solve. Square each binomial.


Standard form
Factor.
Zero-factor property
Solve each equation.

Step 5 State the answer.Since cannot represent a distance, 3 is the only


possible answer. Patricia’s distance is 3 mi, Ali’s distance is


and the distance between them is


Step 6 Check.Since 32 + 42 = 52 ,the answers are correct. NOW TRY


3 + 2 = 5 mi.


3 + 1 =4 mi,


- 1


x= 3 or x=- 1


x- 3 = 0 or x+ 1 = 0


1 x- 321 x+ 12 = 0


x^2 - 2 x- 3 = 0


x^2 +x^2 + 2 x+ 1 =x^2 + 4 x+ 4


In solving a problem involving the Pythagorean theorem, be sure that the ex-


pressions for the sides are properly placed.


1 one leg 22  1 other leg 22 hypotenuse^2


PROBLEM-SOLVING HINT

OBJECTIVE 4 Solve problems by using given quadratic models. In


Examples 1–3,we wrote quadratic equations to model, or mathematically describe,


various situations and then solved the equations. In the last two examples of this sec-


tion, we are given the quadratic models and must use them to determine data.


Step 2 Assign a variable.


Let x Patricia’s distance from the office.


Then Ali’s distance from the office,


and the distance between them.


Place these expressions on a right triangle, as in FIGURE 3.


x+ 2 =


x+ 1 =


=


NOW TRY
EXERCISE 3
The longer leg of a right
triangle is 7 ft longer than the
shorter leg and the hypotenuse
is 8 ft longer than the shorter
leg. Find the lengths of the
sides of the triangle.


NOW TRY ANSWER



  1. 5 ft, 12 ft, 13 ft

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