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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.

 May 8, 2015

Best Answer 

 #4
avatar+27558 
+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
avatar+27558 
+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
avatar+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
avatar+981 
0

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

 May 8, 2015
 #4
avatar+27558 
+13
Best Answer

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
avatar+981 
+5

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

 May 8, 2015
 #6
avatar+99356 
+5

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

 May 8, 2015
 #7
avatar+981 
+5

Thanks :)

 May 8, 2015

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