Question 1031819
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The length of a particular rectangle is 8 inches more than 2 times its width. A new rectangle is formed by tripling 
the width. The area is 30 square inches more than the area of original rectangle. Find the dimensions of the original 
rectangle.
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The solution in the post by @mananth is incorrect due to an arithmetic error on the way.
I came to bring a correct solution.



<pre>
let width be x

length will be 2x+8

Area = x(2x+8)

A new rectangle is formed by tripling the width.

new width = 3x

length = 2x+8

Area = 3x(2x+8)

The area is 30 square inches more than the area of original rectangle.

3x(2x+8) -x(2x+8)=30

2x(2x+8) = 30

4x^2 + 16x - 30 = 0

2x^2 +  8x - 15 = 0


{{{x[1,2]}}} = {{{-2 +- sqrt(46)/2}}}


Taking positive value width = {{{-2 + sqrt(46)/2}}} = 1.39116 inches, approximately.


Length = 2x+8 = 2*1.39116+8 = 10.78232 inches, approximately.
</pre>

Solved.