SOLUTION: The perimeter of a quadrilateral is 160, the sides are (4m^2+1), (3m-4), (2m+6), (4m^2+2m+1) Please help, thanks in advance!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The perimeter of a quadrilateral is 160, the sides are (4m^2+1), (3m-4), (2m+6), (4m^2+2m+1) Please help, thanks in advance!      Log On


   

Question 789771: The perimeter of a quadrilateral is 160, the sides are (4m^2+1), (3m-4), (2m+6), (4m^2+2m+1)
Please help, thanks in advance!
Found 2 solutions by mananth, stanbon:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
(4m^2+1), (3m-4), (2m+6), (4m^2+2m+1)
(4m^2+1)+(3m-4)+ (2m+6)+ (4m^2+2m+1)=160

add up like terms
4m^2+4m^2+3m+2m+2m+1-4+6+1=160
8m^2+7m+4=160
8m^2+7m-156=0
a= 8 , b= 7 , c= -156

b^2-4ac= 49 + 4992
b^2-4ac= 5041
%09sqrt%28%095041%09%29=%0971%09
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( -7 + 71 )/ 16
x1= 4.00
x2=%28-b-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x2=( -7 -71 ) / 16
x2= -4.88
the sides cannot be negative
so x=4
Plug 4 in all the sides to get the lengths


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The perimeter of a quadrilateral is 160, the sides are
(4m^2+1), (3m-4), (2m+6), (4m^2+2m+1)
-----
Add those to get 8m^2+ 7m + 4 = 160
-----
8m^2 +7m - 156 = 0
m = 4
--------------
4m^2+1 = 65
3m-4 = 8
2m+6 = 14
4m^2+2m+1 = 73
==================
Cheers,
Stan H.