Question 249936
This is the only information you need for this
:
The length of the rectangle is 50m more than the width.
 The area of the rectangle is 6496m(squared).
 What are the dimensions of the rectangle?
:
L = W + 50;(length is 50 m more than the width)
;
L * W = 6496; (the area)
:
Replace L with (W+50)
W(W+50) = 6496
:
W^2 + 50W - 6496 = 0; a quadratic equation
:
Use the quadratic formula for this:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
;
In this problem: x=w; b=50; c=-6496
{{{w = (-50 +- sqrt(50^2-4*1*-6496 ))/(2*1) }}}
Do the math here
:
You should get a positive solution
w ~ 59.386 is the width, easy to find length then
Check solution by finding the area