Question 1181876
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I write this response to give hearty endorsement to the method show by tutor @ikleyn for solving this problem.  It is a trick that can make the solution of this and many similar problems easier -- easy enough that someone with good mental math skills can solve it in a very short time.<br>
Let's review the standard algebraic setup for solving the problem.<br>
The perimeter is 120, so the length plus width is 60; and the area is 864.  So we divide the "length plus width" into two parts, x and (60-x).<br>
Multiplying length times width to get the area leads us to the quadratic equation x^2-60x+864=0.<br>
To try to solve that by factoring, we need to find two numbers whose sum is 60 and whose product is 864.<br>
But that's what the original problem required us to do -- so the formal algebra hasn't gotten us any closer to the solution.  We can of course use the quadratic formula to get the solution....<br>
But the other tutor's setup for solving the problem makes the solution far easier.<br>
Instead of breaking the 60 into x and (60-x), break it into (30+x) and (30-x).<br>
That makes the equation (30+x)(30-x)=864, which is easily solved:<br>
900-x^2-864
x^2=36
x=6<br>
The length and width are 30+6=36 and 30-6=24.<br>
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It's a very nice time-saver method for solving any problem which reduces to finding two numbers with a given sum and a given product.<br>
Here is another example, to help you see the method....<br>
Find two numbers whose sum is 50 and whose product is 616.<br>
Let the two numbers be (25+x) and (25-x).  Then<br>
(25+x)(25-x)=616
625-x^2=616
x^2=9
x=3<br>
The two numbers are 25+3=28 and 25-3=22.<br>