Question 222512
If the length of a rectangle is 3 feet longer than the width and the diagonal 
is 5 feet, then what are the length and width?
:
Using a^2 + b^2 = c^2
where
let x = width (a)
then
(x+3) = the length (b)
and
c = 5
:
x^2 + (x+3)^2 = 5^2
FOIL
x^2 + x^2 + 6x + 9 = 25
;
2x^2 + 6x = 9 - 25 = 0
:
2x^2 + 6x - 16 = 0
Simplify, divide by 2
x^2 + 3x - 8 = 0
Use the quadratic formula to find x
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
in this problem: a=1; b=3; c=8
{{{x = (-3 +- sqrt(3^2 - 4 *1*-8 ))/(2*1) }}}
:
Do the math and you should get a positive solution: x=1.7 is the width
and
4.7 is the length
;
;
Check on a calc enter {{{sqrt(1.7^2 + 4.7^2)}}} = 4.998 ~ 5