SOLUTION: please assist me in: 1. solve by completeing the square. p<sup>2</sup> - 8p = 0. 2. solve by graphing x<sup>2</sup> + 6x - 2 = 0

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: please assist me in: 1. solve by completeing the square. p<sup>2</sup> - 8p = 0. 2. solve by graphing x<sup>2</sup> + 6x - 2 = 0       Log On


   



Question 82127This question is from textbook
: please assist me in:
1. solve by completeing the square. p2 - 8p = 0.
2. solve by graphing x2 + 6x - 2 = 0
This question is from textbook

Found 2 solutions by Edwin McCravy, Earlsdon:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

please assist me in: 
1. solve by completing the square. p2 - 8p = 0. 

     p2 - 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:

p2 - 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)2

      (p - 4)2 = 16

Use the principle of square roots:
                   __
         p - 4 = ±Ö16

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.

-----------------------------------

2. solve by graphing x2 + 6x - 2 = 0 

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

y = x2 + 6x - 2
y = 0

Graph each of these by getting some points on each

y = x2 + 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:

 
Now draw a smooth curve through all those points:

 

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:  

 

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



Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
1) Solve bycompleting the square:
p%5E2-8p+=+0 Add the square of half the x-coefficient (%28-8%2F2%29%5E2+=+16) to both sides.
p%5E2-8p%2B16+=+16 Factor the left side.
%28p-4%29%28p-4%29+=+16 or
%28p-4%29%5E2+=+16 Take the square root of both sides.
p-4+=+4 or p-4+=+-4
p+=+8 or p+=+0
2) Solve by graphing:
x%5E2%2B6x-2+=+0
graph%28300%2C200%2C-10%2C5%2C-15%2C5%2Cx%5E2%2B6x-2%29
You can estimate the roots from the graph as:
x = -6.3 and x = 0.3