My Math Forum solve a quadratic equation

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 September 19th, 2011, 12:24 PM #1 Newbie   Joined: Sep 2011 Posts: 13 Thanks: 0 solve a quadratic equation I have this equation, ax^2-(2a+1)x+(a+1)=0 (a>0) should I combine 2a+1 and a+1 then distribute the x? Or if I don't do that, what do I do with the a by itself?
 September 19th, 2011, 12:54 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: solve a quadratic equation We are given: $ax^2-(2a+1)x+(a+1)=0$ where $0. I would just use the quadratic formula: $x=\frac{(2a+1)\pm\sqrt{(2a+1)^2-4(a)(a+1)}}{2a}=\frac{(2a+1)\pm1}{2a}$ Thus: $x=1,\,1+\frac{1}{a}$ Now (observing 1 is a root), we could have factored: $ax^2-ax-(a+1)x+(a+1)=0$ $ax(x-1)-(a+1)(x-1)=0$ $(x-1)$$ax-(a+1)$$=0$ $a(x-1)$$x-(1+\frac{1}{a})$$=0$ Think about what is implied about the roots when the constant a changes in size...

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