Solving Quadratic EquationsDate: 5/23/96 at 10:18:39 From: Anonymous Subject: Solving Quadratic Equations 5x+x(x-7) = 0 Date: 5/23/96 at 10:53:37 From: Doctor Elise Subject: Re: Solving Quadratic Equations Hi! The first thing to do in order to solve any polynomial is to multiply everything out: 5x + x(x-7) = 0 From the distributive property of multiplication, we can say that x(x-7) = (x * x) - (x * 7), which becomes x^2 - 7x 5x + x^2 - 7x = 0 Now we can take the plain old "x" parts and add them: x^2 + 5x - 7x = 0 x^2 - 2x = 0 Okay. Now that we've multiplied everything out and added it together, we have to factor again! Both expressions have an "x", so let's factor out the "x" so that we have: x(x - 2) = 0 So now we have "something" times "something else" equals zero. In order for the equation to equal zero, either the "something" or the "something else" has to equal zero, so there are two possible answers for 'x': if x = 0, we have 0 (0 - 2) = 0, which is true because 0 * -2 does equal 0. if (x - 2) = 0, we have x = 2, so 2 (2 - 2) = 0, which also works. So x = 0 or x = 2 are our possible answers. Hope this helps! -Doctor Elise, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/