Question 1173107
<pre>  
Looks like you're studying "completing the square".

x<sup>2</sup> + 6x + 7 = 0

That won't factor, so we must "MAKE IT FACTOR IN A SPECIAL WAY!!!"

Subtract 7 from both sides, leaving only the two terms with x on the left.

x<sup>2</sup> + 6x     = -7

Now we need to add something to both sides that will make the left side
factor in a SPECIAL way.

Here is the rule:
                    1. Multiply the coefficient of x by 1/2.
                       [That is, 6 times 1/2 gives 3.]
                    2. Square what you get.
                       [We square 3 getting 9.]
                    3. Add that to both sides of the equation.
                       [We add 9 to both sides of the equation:

x<sup>2</sup> + 6x + 9 = -7 + 9

We factor the left side, and combine the numbers on the right.

(x + 3)(x + 3) = 2

Notice that this factored in a SPECIAL way, because both factors
are the same, (x + 3).  So we can write the left side as (x + 3)<sup>2</sup>.

      (x + 3)<sup>2</sup> = 2

Now we can take the square roots of both sides.   Since every number has
two square roots we must annex a ± to the right side when we take the
square roots of both sides:

      {{{sqrt((x+3)^2)}}}{{{""=""}}}{{{"" +- sqrt(2)}}}

When we take the square root of something squared, we just take away the
square root symbol and the square.

       {{{x+3}}}{{{""=""}}}{{{"" +- sqrt(2)}}}

Then we add -3 to both sides of the equation:

        {{{x}}}{{{""=""}}}{{{-3 +- sqrt(2)}}}

So there are two solutions: {{{-3+sqrt(2)}}} and {{{-3-sqrt(2)}}}

Checking the first solution with a calculator:

-3+sqrt(2) = -1.585786438

Substituting that for x

{{{x^2+6x+7}}}{{{""=""}}}{{{0}}}
{{{(-1.585786438)^2+6(-1.585786438)+7}}}{{{""=""}}}{{{0}}}
{{{2.514718626-9.514718628+7}}}{{{""=""}}}{{{0}}}
{{{-2E-9}}}{{{""=""}}}{{{0}}}

That -2E-9 means scientific notation -2 x 10<sup>-9</sup>
which means -0.000000002.

So it means that since all calculators are limited in the number of
decimal places they can handle, we accept that as close enough to 0.

We can check the other solution in the same way.

Edwin</pre>