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```
Date: 9/14/96 at 15:5:59
From: Anonymous

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

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/
```
Associated Topics:
Middle School Algebra

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