a+b+c=0
[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]
[a^2+b^2+c^2+2ab+2ac+2bc=0]
[a^2+b^2+c^2=-(2ab+2ac+2bc)]
[a^2+b^2+c^2=-2(ab+ac+bc)] (i)
also
[a=-b-c]
[a^2=-ab-ac] (ii)
[-c=a+b]
[-bc=ab+b^2] (iii)
adding (ii) and (iii) ,we have
[a^2-bc=b^2-ac] (iv)
devide (i) by (iv)
[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
Answer:
a= 200
b = 210
Step-by-step explanation:
My assumption is, we have to find the length of sides of rectangle
Given
perimeter = 2a + 2b = 820 ft (i) (here a is smaller side and b is larger side)
area = a*b = 42,000 ft^2 (ii)
from eq (1)
2a + 2b = 820
=> 2(a+b) = 820
=> a+b = 820/2
=> a + b = 410
=> a = 410-b (iii)
putting the value of a in eq(ii), we get
(410-b) *b = 42,000
410b - b^2 = 42,000
0 = b^2 - 410b + 42000
b^2 - 410b + 42000 = 0
b^2- 200b- 210b + 42000 = 0
b(b-200)-210(b-200) = 0
(b-200)(b-210) = 0
or
b= 210 and b = 200
if b is larger side than b =210
By putting value of b in eq(iii),
a = 410 -210 = 200
