A farmer needs to enclose a rectangular paddock of Area=130m

[Guest]

A farmer needs to enclose a rectangular paddock of Area=130m^2 with 50m of wiring.
Perimeter = 50m
Area of paddock = 130m²
Here is the Formula I created to solve this:
[tex]l^2-25l+130=0[/tex]
which, when crunched through the Quad Formula, yeilds:
[tex]\frac{25\pm\sqrt{105}}{2}[/tex]
which in turns yields dimensions of paddock to be [tex]\approx 17.6m \& 7.4m[/tex]
My question: how is it that the exponent on 130m^2 can be whisked away like that and still produce a correct answer?

[Jonasmathwork]

Solution: Let x be the width and y be the length in meter
then x+y=50m
xy=130m²
there for y=130m²/x m
y=130m/x
and x=130m²/ym
x=130m/m
there for the value of x and y will be in meter not in meter².

I hope you got your answer.

[Guest]

LB=130 m^2 sq metres and L+B=25 m linear metres(half perimeter), so L=(25-B) m linear metres
B(25-B)=130 equation in sq metres units
B^2 - 25B + 130 = 0 rearranged in standard form
Solve for roots L and B will be linear metres for length and breadth
B = 7.38 or 17.62
L = 7.38 or 17.62 because LB=130 sq metres or L+B=25 metres