Completing the Square and the Quadratic FormulaDate: 09/30/97 at 20:25:45 From: David Subject: Completing the Square I really need help big time on this problem: 1 + 7/x + 2/x^2 = 0 Date: 10/19/97 at 00:11:03 From: Doctor Ezra Subject: Re: Completing the Square Dear David, Thanks for writing in. This looks like a tricky problem, but there's a way to make it easier to see what's going on. Let's try: First, multiply both sides by x^2. That gets rid of the denominators: x^2(1 + 7/x + 2/x^2) = x^2*0 x^2*1 + x^2*7/x + x^2*2/(x^2) = 0 x^2 + 7*x + 2 = 0 If the lefthand side were a perfect square, you could figure out the answer by just taking square roots. But it is not, so we do something called "Completing the Square," which leads to the Quadratic Formula. You can solve the equation ax^2 + bx + c = 0 if the left side is a perfect square: just take the square root. If the left side is not a square, you can change it into one as follows: First, subtract c from both sides and divide by a. You get x^2 + (b/a)x = -c/a. Since (x + n)^2 = x^2 + 2nx + n^2, you can make the left side a perfect square by adding the square of half the coefficient of x to both sides. That number is just (b/2a)^2, or b^2/4a^2. This gives you x^2 + (b/a)x + (b/2a)^2 = b^2/4a^2 -c/a = (b^2 - 4ac)/4a^2 Also, the left side is now the square of x + b/2a, so we have (x + b/2a)^2 = (b^2 - 4ac)/4a^2. Taking square roots, and not forgetting that a nonzero number has two square roots (one being the negative of the other), we get that x + b/2a = +- Sqrt[(b^2 - 4ac)/4a^2], and so x = (-b + Sqrt[b^2 - 4ac])/2a, and x = (-b - Sqrt[b^2 - 4ac])/2a after a bit of algebra. This last expression is the Quadratic Formula, and you can now use it to solve your equation (note that for your equation, a = 1, b = 7, and c = 2). Remember to catch arithmetic errors by substituting the values for "x" that you get via the Quadratic Formula back into your original equation and making sure that you still get zero. Good luck! -Doctor Ezra, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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