SOLUTION: the width of a rectangle is 5 less than twice its lenght, if its area is 3 feet square find the lenght and the width of the rectangle
five less than the square of a positive nu
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-> SOLUTION: the width of a rectangle is 5 less than twice its lenght, if its area is 3 feet square find the lenght and the width of the rectangle
five less than the square of a positive nu
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Question 120099: the width of a rectangle is 5 less than twice its lenght, if its area is 3 feet square find the lenght and the width of the rectangle
five less than the square of a positive number is 1 more than 5 times the number find number Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let l= length of rectangle
Then width (w)=2l-5
Area(a) of a rectangle=length(l) * width(w) so we have:
3=l(2l-5) get rid of parens
3=2l^2-5l subtract 3 from both sides
2l^2-5l-3=0 quadratic in standard form and it can be factored:
(2l+1)(l-3)=0
2l+1=0
l=-1/2------------------no good ---lengths are positive
and
l-3=0
l=3ft-------------length of rectangle
w=2l-5=2*3-5=1ft-----width of rectangle
CK
A=l*w
3=3*1
3=3
Let x=the positive number
x^2-5=five less than the square of the positive#
5x+1=one more than five times the number
so we have:
x^2-5=5x+1 subtract 5x and also 1 from both sides
x^2-5-5x-1=5x+1-5x-1 collect like terms
x^2-5x-6=0 quadratic in standard form and it can be factored:
(x-6)(x+1)=0
x-6=0
x=6-------------------------the positive number
and
x+1=0
x=-1------------------negative number---disregard
x
CK
x^2-5=5x+1
6^2-5=5*6+1
36-5=30+1
31=31
