Question 82127
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please assist me in: 
1. solve by completing the square. p<sup>2</sup> - 8p = 0. 

     p<sup>2</sup> - 8p = 0

Multiply the coefficient of p, which is -8 by 1/2.
That gives -4.  Square -4. That gives +16. Add 16
to both sides of the equation:

p<sup>2</sup> - 8p + 16 = 0 + 16

Factor the left side. The right side is just 16

(p - 4)(p - 4) = 16

The left side can be written (p - 4)<sup>2</sup>

      (p - 4)<sup>2</sup> = 16

Use the principle of square roots:
                   __
         p - 4 = ±<font face = "symbol">Ö</font>16

16 is the square of 4, so 4 is the positive square root of 16

         p - 4 = ±4

Add 4 to both sides:

             p = 4 ± 4

Using the +, p = 4 + 4 = 8

Using the -, p = 4 - 4 = 0

The solutions are 8 and 0.

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2. solve by graphing x<sup>2</sup> + 6x - 2 = 0 

Write this as the system of equations by
letting y = the left side and y = the right side:

y = x<sup>2</sup> + 6x - 2
y = 0

Graph each of these by getting some points on each

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

Some points are (-7,5), (-6,-2), (-5,-7), (-4,-10), 
(-3, -11), (-2,-10), (-1,-7), (0,-2), (1,5)

Plotting these:

{{{drawing(244.4,400, -8, 2, -12, 6, 

   locate(-7.1,5.4,o), locate(-6.15,-1.6,o), locate(-5.1,-6.6,o), locate(-4.16,-9.5,o),
   locate(-3.2,-10.6,o), locate(-2.12,-9.5,o), locate(-1.1,-6.5,o), locate(-.15,-1.6,o), 
   locate(.85,5.35,o),    

    graph(244.4, 400, -8, 2, -12, 6) )}}} 
Now draw a smooth curve through all those points:

{{{drawing(244.4,400, -8, 2, -12, 6, 

   locate(-7.1,5.4,o), locate(-6.15,-1.6,o), locate(-5.1,-6.6,o), locate(-4.16,-9.5,o),
   locate(-3.2,-10.6,o), locate(-2.12,-9.5,o), locate(-1.1,-6.5,o), locate(-.15,-1.6,o), 
   locate(.85,5.35,o),    

    graph(244.4, 400, -8, 2, -12, 6, x^2+6x-2) )}}} 

Now we draw the equation of y = 0, but this is just the x-axis
so we want to find the points where the curve crosses the x-axis.

The best we can do is estimate what we think these points are.

I'll mark these on the x-axis and what I think they are:  

{{{drawing(244.4,400, -8, 2, -12, 6, 

   locate(-7.1,5.4,o), locate(-6.15,-1.6,o), locate(-5.1,-6.6,o), locate(-4.16,-9.5,o),
   locate(-3.2,-10.6,o), locate(-2.12,-9.5,o), locate(-1.1,-6.5,o), locate(-.15,-1.6,o), 
   locate(.85,5.35,o), locate(.22,.4,O), locate(-6.45,.4,O), 

   locate(1,1.5,.3), locate( .6,1,"/"), locate(-6,1.5,-6.3), locate( -6,1,"/"),     

    graph(244.4, 400, -8, 2, -12, 6, x^2+6x-2) )}}} 

I "guess"-timate that the one on the right is about
1/3 of the way between 0 and 1, so I'll call it .3.

I slso "guess"-timate that the one on the left is about
1/3 of the way between -6 and -7, so I'll call it -6.3.

Edwin</pre>