The length of a room is 1 more than 3 times its width. The area of the room is 80 square meters. Find the demensions.
Let L be the length and W the width
L = 3W + 1 (1) (I assume the room is 1 metre longer than 3 times the width)
L*W = 80 (2)
Using (1) in (2)
(3W + 1)W = 80
Rearrange
3W2 + W - 80 = 0
This factors as
(3W + 16)(W - 5) = 0
Since we can't have a negative width for the room, the only valid solution is W = 5m
Using (1) this means that L = 3*5 + 1 = 16m
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The length of a room is 1 more than 3 times its width. The area of the room is 80 square meters. Find the demensions.
$$w\times{l}=A$$
$$w\times({3w+1})=80$$
$$3w^2+w=80$$
$$3w^2+w-80=0$$
$$w=\frac{-1+\sqrt{1^2-4\times{3}\times{-80}}}{2\times{3}}$$ or $$w=\frac{-1-\sqrt{1^2-4\times{3}\times{-80}}}{2\times{3}}$$
$$w=\frac{-1+\sqrt{961}}{6}$$ or $$w=\frac{-1-\sqrt{961}}{6}$$
$$w=\frac{{-1}+31}{6}$$
$$\mathbf{w=5}$$
$$l=3\times{5}+1}$$
$$\mathbf{l=16}$$
Fixed!
You are nearly right zacismyname. However, you should take a closer look at your calculation of $$1^2-4\times3\times-80$$ under the square root sign. It isn't 960 (almost, but not quite!).
You've nothing to be ashamed of!
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Bad choice of word I would have just preferred to factorise rather than use the formula.