Solving PolynomialsDate: 6/16/96 at 17:50:5 From: Anonymous Subject: Solving polynomials Please help me!!! My final exam is tomorrow morning! x^2 + 2x + 3 26 -------- -------- = ---- 3 x^2 + 2x 5 I've tried doing this by finding a common denominator, but after that everything gets confusing and starts looking like this: 5x^4 + 20x^3 - 58x^2 - 156x + 45 -------------------------------- = 0 15x^2 + 30 Thanks for your help! Lisa Date: 6/16/96 at 19:2:25 From: Doctor Anthony Subject: Re: Solving polynomials You could continue and simply put the numerator of above expression equal to zero and solve the quartic, but that would be extremely tedious unless you have Maple or a TI-92. I expect you went through unnecessary steps in deriving the above equation, but I would not do it the way you started out. Look at the expression again, and you will see that there is a similarity of the terms on the left hand side. The second term is just the reciprocal of the first term. I am going to let y = x^2+2x, and so we get: y/3 + 3/y = 26/5 or (y^2+9)/3y = 26/5 Multiply out 5y^2 + 45 = 78y 5y^2 - 78y + 45 = 0 Factorize (5y-3)(y-15) = 0 and so y = 3/5 or y = 15 Now go back to the x variable and we have: x^2 + 2x = 3/5 and x^2 + 2x = 15 Solve these two quadratics to get the four x values that you need. The second quadratic is x^2 + 2x - 15 = 0 (x+5)(x-3) = 0 so -5 and 3 will be two of the solutions. The other quadratic doesn't factorize, so use the quadratic formula to solve 5x^2 + 10x - 3 = 0 I get x = -2.264911 and 0.264911 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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