Question 628329
The length of a rectangle is 4 meters longer than the width.
 If the area is 29 square meters, find the rectangles dimensions.
 Round to the nearest tenth of a meter. 
:
Let W = the width
It says,"The length is 4 meters longer than the width", therefore
L = W+4
:
The area = 29 sq/meters, therefore;
L * W = 29
Replace L with (W+4)
(W+4)*W = 29
W^2 + 4W = 29
A quadratic equation
W^2 + 4W - 29 = 0
Use the quadratic formula to find W
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem: x=W, a=1, b=4, c=-29
{{{W = (-4 +- sqrt(4^2-4*1*-29 ))/(2*1) }}}
:
{{{W = (-4 +- sqrt(16-(-116) ))/2 }}}
:
{{{W = (-4 +- sqrt(132 ))/2 }}}
Two solutions, we only want the positive solution
{{{W = (-4 + 11.489)/(2*1) }}}
W = {{{7.489/2}}}
W ~ 3.7 meters is the width
then
3.7 + 4 ~ 7.7 is the length
:
:
Check this by finding the area: 
7.7 * 3.7 = 28.49, not 29 because rounded to nearest 10th