Answer The value of x = 10. The dimensions of the enclosure is, Explanation Given: It is given that a rectangular enclosure at the zoo is 35 feet long by 18 feet wide. The zoo wants to double the area of the enclosure by adding the same distance x to the length and width. The diagram shown is related to given information, Here, Area of the rectangle = = l e n g t h × b r e a d t h = 18 × 35 = 630 f e e t 2 Now, Zoo wants to double the area, Then, So, Area of rectangle = l e n g t h × b r e a d t h 1260 = ( 18 + x ) ( 35 + x ) 18 ( 35 + x ) + x ( 35 + x ) = 1260 630 + 18 x + 35 x + x 2 = 1260 x 2 + 53 x − 630 = 0 Now, Solving the above quadratic equation , x 2 + 53 x − 630 = 0 x 2 + 63 x − 10 x − 630 = 0 x ( x + 63 ) − 10 ( x + 63 ) = 0 ( x − 10 ) ( x + 63 ) = 0 x = 10 , − 63 ‘ x’ cannot be negative. So, the value of x = 10. The dimensions of the enclosure is,

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

1st Edition

ISBN: 9781680331165

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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Question

Chapter 3, Problem 5CR

To determine

To evaluate: the solution of an equation and the dimensions of the enclosure.

Expert Solution & Answer

Answer to Problem 5CR

The value of x = 10.

The dimensions of the enclosure is,

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015, Chapter 3, Problem 5CR , additional homework tip 1

Explanation of Solution

Given:

It is given that a rectangular enclosure at the zoo is 35 feet long by 18 feet wide. The zoo wants to double the area of the enclosure by adding the same distance x to the length and width.

The diagram shown is related to given information,

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015, Chapter 3, Problem 5CR , additional homework tip 2

Here,

Area of the rectangle =

=length×breadth=18×35=630feet2

Now,

Zoo wants to double the area,

Then,

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015, Chapter 3, Problem 5CR , additional homework tip 3

So,

Area of rectangle = length×breadth

1260=(18+x)(35+x)18(35+x)+x(35+x)=1260630+18x+35x+x2=1260x2+53x−630=0

Now,

Solving the above quadratic equation,

x2+53x−630=0x2+63x−10x−630=0x(x+63)−10(x+63)=0(x−10)(x+63)=0x=10,−63

‘ x’ cannot be negative.

So, the value of x = 10.

The dimensions of the enclosure is,

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015, Chapter 3, Problem 5CR , additional homework tip 4

Chapter 3, Problem 4CR

Chapter 3, Problem 6CR

Chapter 3 Solutions

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

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