Question 1124076
<br>
Let x be the width; then the length is x+5.  Then the area is length times width:<br>
{{{x(x+5) = 204}}}
{{{x^2+5x-204 = 0}}}
{{{x+17)(x-12) = 0}}}
{{{x = -17}}} or {{{x = 12}}}<br>
Obviously -17 is not a valid width.  So the width is 12m and the length is 17m.<br>
To solve that problem algebraically, as above, we had to factor the x^2+5x-204 by finding two number whose product is 204 and whose difference is 5.<br>
But that is exactly what the problem asks us to do... so the formal algebra doesn't help us any.<br>
The easiest way to solve this problem is to forget the algebra and straight away look for two numbers whose product is 204 and whose difference is 5.  A little trial and error finds the numbers 12 and 17.