SOLUTION: A rectangle is 2 inches longer than it is wide. Numerically, its area exceeds its perimeter by 31. Find the perimeter.

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Question 1170327: A rectangle is 2 inches longer than it is wide. Numerically, its area exceeds its perimeter by 31. Find the perimeter.

Found 3 solutions by josgarithmetic, ikleyn, Clanther:
Answer by josgarithmetic(39618) About Me  (Show Source):
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dimensions n+2 and n

perimeter 2%28n%2B2%29%2B2n=2n%2B4%2B2n=4n%2B4

area n%5E2%2B2n

Condition: n%5E2%2B2-4n-4=31
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Answer by ikleyn(52797) About Me  (Show Source):
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.

Let x be the width, in inches;

then the length is (x+2) inches.


The area is  x*(x+2) square inches.


The perimeter is  2x + 2*(x+2) = 4x + 4 inches.


The equation is  

    x*(x+2) - (4x+4) = 31,   or


    x^2 +2x - 4x + 2 - 4 = 31

    x^2 - 2x - 35 = 0

    (x-7)*(x+5) = 0.


Of the two roots  x= 7, x= -5, only value of 7 is positive and is the solution to the problem.


The dimensions of the rectangle are  7 and 9 inches.      ANSWER


CHECK.  The area is 7*9 = 63 sq.inches.  The perimeter is  2*(7+9) = 32 inches.  The difference is  63 - 32 = 31.   ! Correct !

Solved.



Answer by Clanther(6) About Me  (Show Source):
You can put this solution on YOUR website!
let x be the width
let x+2 be the length
A=area
P=perimeter
A=lw
=x(x+2)
A=x²+2x
P=2x+2(x+2)
=2x+2x+4
P=4x+4
The equation is:
area-perimeter=31
(x²+2x)-(4x+4)=31
x²+2x-4x-4=31
x²-2x-4-31=0
x²-2x-35=0
by completing the square:
x²-2x+1=35+1
(x-1)²=36
√(x-1)²=±√36
x-1=±6
Take the positive value of the roots/factors
x=1+6
x=7 - the width
Substitute the value of x to the length representation:
x+2= 7+2= 9-the length
Perimeter: Substitute the positive value of the root(width) to:
P=4x+4
=4(7)+4
P=32"