Solving quadratic equations using the method from India

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DD2162

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The equation is x^2 + 12x - 64 = 0. I think I have the first three steps right:
x^2 + 12x = 0 + 64
4 (x^2) + 4(12x) = 4 (64) or 4x^2 + 48x = 256
then I squared the coefficient of the original x term and added it to both sides
4x^2 + 48x + 12^2 = 256 + 12^2 or 4x^2 + 48x + 144 = 256 + 144
from here I am stuck. I do not understand how to find the square root of both sides
nor do I get the last two steps. Can you please help?
 
DD2162 said:
The equation is x^2 + 12x - 64 = 0. I think I have the first three steps right:
x^2 + 12x = 0 + 64
4 (x^2) + 4(12x) = 4 (64) or 4x^2 + 48x = 256
then I squared the coefficient of the original x term and added it to both sides
4x^2 + 48x + 12^2 = 256 + 12^2 or 4x^2 + 48x + 144 = 256 + 144
from here I am stuck. I do not understand how to find the square root of both sides
nor do I get the last two steps. Can you please help?
Click to expand...
Edit: I have revised this post to take into account lookagain's post about a method of solving quadratics previously not known to me.

There are four ways (at least) to solve a quadratic equation.

The one you seem to be being taught is this:

\(\displaystyle x^2 + 12x - 64 = 0 \implies x^2 + 12x = 64.\)

\(\displaystyle x^2 + 12x = 64 \implies (4 * 1)(x^2 + 12x) = (4 * 1)64 \implies 4x^2 + 48x = 256 \)

All this is straight forward. This method now requires you to add the square of the original coefficient of x to both sides of the equation.

The original coefficient is 12 so its square is 144.

\(\displaystyle 4x^2 + 48x = 256 \implies 4x^2 + 48x + 144 = 256 + 144 = 400 \implies 4(x^2 + 12x + 36) = 20^2 \implies 2^2(x + 6)^2 = 20^2.\)

Now wrap it up.
 
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DD2162 said:
4x^2 + 48x + 144 = 256 + 144

from here I am stuck

I do not understand how to find the square root of both sides
Click to expand...

At this point, I would simplify and factor. (I don't know what the Indian method would do.)

x^2 + 12x + 36 = 100

If you already understand that 6*6=36 and 6+6=12, then you have (x+6)(x+6) on the left side:

(x+6)^2 = 100

Can you take the square root of both sides and finish?
 
DD2162 & JeffM edit said:
The equation is x^2 + 12x - 64 = 0. I think I have the first three steps right:
x^2 + 12x = 0 + 64 Fine
4 (x^2) + 4(12x) = 4 (64) or 4x^2 + 48x = 256 Why are you multiplying by 4? *
then I squared the coefficient of the original x term and added it to both sides
4x^2 + 48x + 12^2 = 256 + 12^2 or 4x^2 + 48x + 144 = 256 + 144
from here I am stuck. I do not understand how to find the square root of both sides
nor do I get the last two steps. Can you please help?
Click to expand...

* See: http://www.algebra.com/algebra/homework/quadratic/THEO-2011-08-28-02.lesson <------- "Lesson THE INDIAN METHOD OF SOLVING A QUADRATIC EQUATION"
 
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