Solve by FactoringDate: 13 Mar 1995 11:29:24 -0500 From: Richard Seguin Subject: your mail Hello I had a question in math class like this today: 3w^2 + 40w - 25 = 20 - 2w^2 We are doing a section On Solving Equations by Factoring. Could you help me? Date: 13 Mar 1995 21:53:19 -0500 From: Dr. Sydney Subject: Re: your mail Hello again, Richard! Okay, for your problem, our final goal is to find what values of w make the equation hold true. The way we usually do this is to get everything on one side of the equation, so we have an expression with w's and numbers on the left, and 0 on the right. Here is what I mean: Subtract 20 and add 2w^2 to both sides of the equation. Then you get: 3w^2 + 40w - 25 - 20 + 2w^2 = 0 Combine the w^2 terms with one another, and do the same with the constant terms to get: 5w^2 + 40w - 45 = 0 Now, do you see anything that factors out of the left hand side? All of the terms on the left are divisible by 5, so let's divide each side of the equation by 5. We then get: w^2 + 8w - 9 = 0 This is much simpler to work with, yes? Okay, now we simply need to factor this quadratic. I always do this by playing around with the numbers... We know that if this factors nicely, it will factor into two terms that look like this: (w + a)(w + c), where a and c are constants. Now, since the constant term is -9 we want ac to be -9. And, since the term with the w has a coefficient of 8, we know that we want a + c = 8. (this is because (w+a)(w+c) = w^2 + (a+c)w + ac). You can figure out what a and c are by using these two equations, or you can simply try a few numbers and see if you can find something that will work. See if you can figure out what a and c are... Can you factor this one? Write back if you want to check your answer, or if you have any more questions! --Sydney, "Dr. Maaaaath" |
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