Question 299944
before u go through all this, pls read the last 4 lines
area=80cm2,
to make things easier, lets call length 'b' and width 'w'
{{{w=(b/2)+3}}}
area=b*w=80
substituting 'w' with {{{(b/2)+3}}}
{{{b*[(b/2)+3]=80}}}
{{{b^2/2 +3b=80}}}
multiply through by 2 to get rid of the fraction
{{{b^2+6b=160}}}
now i will equate everything to zero
{{{b^2+6b-160=0}}}      ......(i)
multiply the first figure(the one with the square to the number without any letter(dont forget the signs)in this case....
{{{b^2 * -160 =-160b^2}}}
now this is the tricky part. from (i) u need to find 2 numbers whose multiple gives you {{{-160b^2}}} and whose sum gives you 6b.
in this case, its 16b and -10b
now all we have to do is replace 6b with this sum, after all, its all the same
      ****dont give up, this only looks bulky cos im tryin to explain clearly***
{{{b^2+16b-10b-160=0}}}
watch closely. this is how to contract. it doesn't matter how its arranged, just take the first 2 no.s and use what they have in common and do the next 2...
{{{b^2+16b-10b-160=0}}}
b(b+16)-10(b+16)=0
(b+16)(b-10)=0
b+16=0                       or       b-10=0
b=-16  (not possible)                 b=10
therefore, b=10
....this might be your problem....were looking for w, not b
w=(b/2)+3
w=(10/2)+3
w=5+3 =8
voila!