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The area of a square can be tripled by increasing its length by 6 centimeters and increasing its width by 3 centimeters. What is the length of the side of the square?

Whiz333  Apr 15, 2017

Best Answer 

 #1
avatar+4777 
+2

We can make these equations from the info in the question:

a = s * s

3a = (s + 6) * (s + 3)

 

Substitute.

3(s * s) = (s + 6) * (s + 3)

 

3s2 = s2 + 9s + 18

 

Subtract 3s2 from both sides.

0 = -2s2 + 9s + 18

 

Now we can use the quadratic formula to solve for s.

\(s = {-9 \pm \sqrt{9^2-4(-2)(18)} \over 2(-2)} \\~\\ s= {-9 \pm \sqrt{81+144} \over -4} \\~\\ s= {-9 \pm \sqrt{225} \over -4} \\~\\ s= \frac{-9\pm15}{-4} \\~\\s =\frac{ -9 + 15} { -4}=-\frac32\text{ or } s=\frac{-9-15}{-4}=6\)

 

We know the side length is positive.

\(s=6\text{ cm}\)

hectictar  Apr 15, 2017
edited by hectictar  Apr 15, 2017
edited by hectictar  Apr 15, 2017
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3+0 Answers

 #1
avatar+4777 
+2
Best Answer

We can make these equations from the info in the question:

a = s * s

3a = (s + 6) * (s + 3)

 

Substitute.

3(s * s) = (s + 6) * (s + 3)

 

3s2 = s2 + 9s + 18

 

Subtract 3s2 from both sides.

0 = -2s2 + 9s + 18

 

Now we can use the quadratic formula to solve for s.

\(s = {-9 \pm \sqrt{9^2-4(-2)(18)} \over 2(-2)} \\~\\ s= {-9 \pm \sqrt{81+144} \over -4} \\~\\ s= {-9 \pm \sqrt{225} \over -4} \\~\\ s= \frac{-9\pm15}{-4} \\~\\s =\frac{ -9 + 15} { -4}=-\frac32\text{ or } s=\frac{-9-15}{-4}=6\)

 

We know the side length is positive.

\(s=6\text{ cm}\)

hectictar  Apr 15, 2017
edited by hectictar  Apr 15, 2017
edited by hectictar  Apr 15, 2017
 #2
avatar
0

Please explain: How can a SQUARE have a side of 6cm and 9cm? Also, how can the AREA increase from a to 6a, when the area is supposed to "triple"?? Thanks.

Guest Apr 15, 2017
 #3
avatar+4777 
0

Woops! You're right... I don't know how I misread that! It should be 3a, not 6a....

 

 

And I figure the new shape is a rectangle, and the old shape is a square.

hectictar  Apr 15, 2017

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