Completing the Square to Solve Two Equations

Date: 5/30/96 at 3:40:15
From: Anonymous
Subject: Complete the square

Dear Dr Math:

I have these two math problems: 
  
  1) x^2-4x-10 = 0
  2) 3x^2-6x-4 = 0

I have to solve these equations by completing the square. Can you 
help?

Thanks.

Date: 5/30/96 at 11:43:22
From: Doctor Anthony
Subject: Re: Complete the square

1) To complete the square, first check that the coefficient of x^2 
is 1. If it is not 1, divide through by the coefficient. In our first 
problem it is 1, so no division is necessary. 

Next look at the coefficient of x. In the first problem it is -4. 
So now take 4 and halve it (=2) and square the result. This gets us 
back to 4 again.  Add 4 as the term required to complete the square. 
Remember to subtract it again to keep the original equation unchanged. 
We get: 

       (x^2 - 4x + 4) - 4 - 10 = 0
                  (x-2)^2 - 14 = 0
                       (x-2)^2 = 14

Now take the square root of both sides

              (x-2) = + or -sqrt(14)
                  x = 2 + or -sqrt(14)

The two roots are x = 2+sqrt(14)  and  x = 2-sqrt(14)    

2) In the second problem we must first divide through by 3 to make the 
coefficient of x^2 equal to 1. This gives x^2 - 2x - 4/3 = 0
Now complete the square by taking coefficient of x (=-2), halve it    
(=-1) and square it (=1).  So write the equation as:

           (x^2 - 2x + 1) -1 - 4/3 = 0
                     (x-1)^2 - 7/3 = 0
                           (x-1)^2 = 7/3

Take the square roots of both sides:

               x-1 = +or-sqrt(7/3)
                 x = 1 +or-sqrt(7/3)

The two roots are x = 1+sqrt(7/3)   and  x = 1-sqrt(7/3)  

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/