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Question 151040This question is from textbook
: If AB=(4n+1)cm, AD=(n-5)cm, and the area of rectangle ABCD is 25cm squared, find AB and AD.This question is from textbook
: If AB=(4n+1)cm, AD=(n-5)cm, and the area of rectangle ABCD is 25cm squared, find AB and AD.
Answer by ptaylor(1256) About Me  (Show Source):
You can put this solution on YOUR website!
Area of Rectangle =Length(L) X Width(W) or A=LW
In this case, let L=AB=4n+1 and let W=AD=n-5, so our equation to solve is:
(4n+1)(n-5)=25 ; Expand using FOIL
4n^2-20n+n-5=25 subtract 25 from each side and collect like terms
4n^2-19n-30=0 quadratic in standard form. Solve using the quadratic formula
n = (-b +- sqrt( b^2-4*a*c ))/(2*a)
n = (19 +- sqrt( (-19)^2+4*4*30 ))/(2*4)
n= (19 +- sqrt( 361+480 ))/(8)
n = (19 +- 29)/8
n=(19+29)/8=48/8=6
and
x=(19-29)/8=-10/8
CK for n=6
AB=4n+1=4*6+1=25 cm-------------------------AB
AD=n-5=6-5=1 cm----------------------------------AD
and AB*AD=25*1=25 good!
CK for n=-10/8
4n+1=4*(-10/8)+1=-5+1=-4 cm-----------No good!! Lengths in this case are positive
Hope this helps---ptaylor