+0

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.

0
722
7

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.

May 8, 2015

#4
+28029
+13

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!

.

May 8, 2015

#1
+28029
+8

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

.

May 8, 2015
#2
+981
+5

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!

May 8, 2015
#3
+981
0

. . . I'm a bit ashamed I didn't see those factors

May 8, 2015
#4
+28029
+13

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!

.

Alan May 8, 2015
#5
+981
+5

Bad choice of word I would have just preferred to factorise rather than use the formula.

May 8, 2015
#6
+101771
+5

You make a great contribution to this forum Zac :))

May 8, 2015
#7
+981
+5

Thanks :)

May 8, 2015