A field with an area 300 square meters has a length that is 5 more meters more then its width. If the width of the field is w meters, then an equation that can be used to determine the value of w is:
a) w^2-1500=0
b) w^2-295=0
c) w^2+5w-300=0
d) w^2-5w-300=0
e) 5w^2-300=0
could you please include a step-by-step explanation??
a) w^2-1500=0
b) w^2-295=0
c) w^2+5w-300=0
d) w^2-5w-300=0
e) 5w^2-300=0
could you please include a step-by-step explanation??
-
The answer is c)
What we know:
the area, A = 300
the width = w
the lenght, l = w + 5
so, width x length = area
therefore w(w + 5) = 300
so expanding the brackets: w^2 + 5w = 300
which is the same as w^2 + 5w - 300 = 0
To solve the equation, factorise the equation:
(w + 20)(w - 15) = 0
so w + 20 = 0 or w - 15 = 0
so w = -20 or w = 15
The width can't be a negative value so it is 15m
The length is w + 5 = 15 + 5 = 20m
20m x 15m = 300m^2 so the values add up to what we expect
What we know:
the area, A = 300
the width = w
the lenght, l = w + 5
so, width x length = area
therefore w(w + 5) = 300
so expanding the brackets: w^2 + 5w = 300
which is the same as w^2 + 5w - 300 = 0
To solve the equation, factorise the equation:
(w + 20)(w - 15) = 0
so w + 20 = 0 or w - 15 = 0
so w = -20 or w = 15
The width can't be a negative value so it is 15m
The length is w + 5 = 15 + 5 = 20m
20m x 15m = 300m^2 so the values add up to what we expect
-
The area is the product of length times width. 300 sq meters area is given so: 300=(W+5)timesW. So that is the formula, now you do the arithmetic.