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Solving Quadratic Equations

Date: 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


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
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Associated Topics:
High School Basic Algebra
Middle School Algebra

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