Question 23410: 10c2-21c=-4c+6
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! 
Are you sure you didn't copy this one wrong? It works out a LOT better (it factors!) if you had written this  . I'm going to take the liberty of changing it, and if I'm wrong, then just repost the question. I will owe you a solution to the other problem, using the quadratic formula (or one of algebra.com quadratic equation solver!)
So I'm saying let's do this:
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This is a quadratic equation, so set it equal to zero, by adding +4c +6 to each side:
This can be a tough factoring problem. You might want to see a page on my website: click on my tutor name "rapaljer" anywhere in algebra.com, and look for "Basic Algebra", then go to "Samples of Basic Algebra: One Step at a Time." Then in Chapter 2 there are several sections on factoring. The one you need here for this problem is the one I call "Advanced Trinomial Factoring." It is a DETAILED explanation, with examples, exercises, and solutions to ALL the exercises-- and like everything else around here, it's ALL FREE!!
Back to the problem:
You need to find two numbers whose product is 10c^2. Most likely that will be 5c*2c:
Now, you need to find two numbers whose product is +6, and the middle term has to add up to a -17c. The signs will both be negative, so try -6*-1 (since I happen to have already tried the -3*-2 and it didn't work out right. I think you should putting the -6 first and the -1 in the second slot.
If you will do the OUTER times OUTER and the INNER times INNER, you will have -5c -12c, which is -17c.
Now finish the problem by separating the quadratic equation into TWO solutions:
First solution:
Second solution:
R^2 at SCC
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