Solving Quadratic EquationsDate: 9/14/96 at 15:5:59 From: Anonymous Subject: Solving Quadratic Equation How do you solve 3x^2 + x - 14 = 0 And another one: x^2 - 6x + 8 = 0 Date: 9/16/96 at 15:18:13 From: Doctor Mike Subject: Re: Solving Quadratic Equation Hello, Each equation is in the form Ax^2 + Bx + C = 0 , so the Quadratic Formula says the solutions are (-B+R)/(2A) and (-B-R)/(2A) where R = sqrt(B^2 - 4*A*C) . Is that familiar? Using this, the 2 solutions for the first equation are 2 and -7/3 . The 2 solutions for the second equation are 2 and 4 . So, 2 is the only solution for both equations. If you can factor these 2 expressions, then that will be another way to show this. In factored form the 2 equations are (3x+7)*(x-2)=0 and (x-4)*(x-2)=0 . This shows that x-2 is the common factor of both expressions, so that the value of x making x-2 = 0 true will be the common solution for the 2 equations. I hope this helps. -Doctor Mike, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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