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### Completing the Square and the Quadratic Formula

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Date: 09/30/97 at 20:25:45
From: David
Subject: Completing the Square

I really need help big time on this problem:

1 + 7/x + 2/x^2 = 0
```

```
Date: 10/19/97 at 00:11:03
From: Doctor Ezra
Subject: Re: Completing the Square

Dear David,

Thanks for writing in. This looks like a tricky problem, but there's a
way to make it easier to see what's going on. Let's try:

First, multiply both sides by x^2. That gets rid of the denominators:

x^2(1 + 7/x + 2/x^2) = x^2*0
x^2*1 + x^2*7/x + x^2*2/(x^2) = 0
x^2 + 7*x + 2        = 0

If the lefthand side were a perfect square, you could figure out the
answer by just taking square roots. But it is not, so we do something

You can solve the equation ax^2 + bx + c = 0 if the left side is a
perfect square: just take the square root. If the left side is not a
square, you can change it into one as follows:

First, subtract c from both sides and divide by a. You get

x^2 + (b/a)x = -c/a.

Since (x + n)^2 = x^2 + 2nx + n^2, you can make the left side a
perfect square by adding the square of half the coefficient of x to
both sides. That number is just (b/2a)^2, or b^2/4a^2. This gives you

x^2 + (b/a)x + (b/2a)^2 = b^2/4a^2 -c/a
= (b^2 - 4ac)/4a^2

Also, the left side is now the square of x + b/2a, so we have

(x + b/2a)^2 = (b^2 - 4ac)/4a^2.

Taking square roots, and not forgetting that a nonzero number has two
square roots (one being the negative of the other), we get that

x + b/2a = +- Sqrt[(b^2 - 4ac)/4a^2], and so
x = (-b + Sqrt[b^2 - 4ac])/2a, and
x = (-b - Sqrt[b^2 - 4ac])/2a

after a bit of algebra. This last expression is the Quadratic Formula,
and you can now use it to solve your equation (note that for your
equation, a = 1, b = 7, and c = 2).

Remember to catch arithmetic errors by substituting the values for "x"
equation and making sure that you still get zero.

Good luck!

-Doctor Ezra,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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