Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

  
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

774,965 users

A rectangle has an area of x2+6x+8

0 votes

 A rectangle has an area of x2+6x+8 the length is x+4. Find the width of the rectangle. could the rectangle be a square? explain why or why?I need help! Please

asked Apr 21, 2014 in ALGEBRA 2 by anonymous

1 Answer

0 votes

Given :

The area of a rectangle(A ) = x ^2 + 6x + 8.

The length of the rectangle(l ) = x + 4.

Area of a rectangle(A ) = length * width (l  * w).

Substitute the values of A = x ^2 + 6x + 8 and l = x + 4.

x ^2 + 6x + 8 = (x + 4) * w

factor x ^2 + 6x + 8 by factor by grouping.

x ^2 + 6x + 8 = x ^2 + 4x + 2x + 8

                   = x (x + 4) + 2(x+ 4)

                   = (x + 4)(x + 2).

(x + 4)(x + 2) = (x + 4) * w

w = [ (x + 4)(x + 2) ] / (x + 4)

w = x + 2.

Thus, width of the rectangle is x + 2.

Let us assume that, the rectangle be a square,

Then, length = width.

x + 4 = x + 2

4 = 2.

The above statement is false, so we can' t assume that, the rectanglr be a square.

answered Apr 21, 2014 by lilly Expert

Related questions

...