SOLUTION: x2+12x-5=0 that is x squared+12x-5=0 this must be solved by the method for solving quadratis equation that came from India, that states- it breaks down like this, a) move the c

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Question 675129: x2+12x-5=0
that is x squared+12x-5=0
this must be solved by the method for solving quadratis equation that came from India, that states-
it breaks down like this, a) move the constant term to the right side of the equation, b) multiply each term in the equation by four times the coefficient of the X squared term, c) squared the coefficient of the original X term and add it to both sides of the equation, D) take the square root of both sides, E) set the left side of the equation equal to the positive square root of the number on the right side and solve for x, F) set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for X
Thank you for your help and please show your work so I can understand how to do this. Thank you.
Found 2 solutions by ewatrrr, Edwin McCravy:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

x2+12x-5=0

x2+12x = 5

4x2+48x = 20

4x2+48x+144 = 164 

sqrt%28%282x%2B12%29%282x%2B12%29%29 = ± sqrt(164)

      2x + 12 = 2sqrt(41)

           x = -6 +sqrt(41)

      2x + 12 = -2sqrt(41)

            x = -6 -sqrt(41)



Checking using 'Completing the Square'

(x+6)^2 = 5 + 36

   x+6 =  ±sqrt(41)

    x = -6 ±sqrt(41)

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
      x² + 12x - 5 = 0

a) move the constant term to the right side of the equation, 
          x² + 12x = 5

b) multiply each term in the equation by four times the coefficient of the X squared term, 
         4x² + 48x = 20

c) square the coefficient of the original X term and add it to both sides of the equation,
   4x² + 48x + 144 = 20 + 144
 
   4x² + 48x + 144 = 164

D) take the square root of both sides,
           2x + 12 = ±√164

           2x + 12 = ±2√41

E) set the left side of the equation equal to the positive square root of the number on the right side and solve for x,
 
          2x + 12 = 2√41

               2x = -12 + 2√41

               2x%2F2 = %28-12%29%2F2 + 2sqrt%2841%29%2F2

                x = -6 + √41

F) set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for X
          2x + 12 = -2√41

               2x = -12 - 2√41

               2x%2F2 = %28-12%29%2F2 - 2sqrt%2841%29%2F2

                x = -6 - √41                


Comment:  See the lesson on this site:

http://www.algebra.com/algebra/homework/quadratic/THEO-2011-08-28-02.lesson

The Indian method has the advantage of avoiding fractions when the coefficient
of x is an integer not evenly divisible by the coefficient of x².  Since your
quadatic equation has 1 as the coefficient of the x² term, the coefficient of x
is divisible by 1 so using the Indian method has no advantage. 

Edwin