Question 373661: I need to solve these quadratic equations using the method that came from India. I am confused.
The sample problem is:
xsquared + 3x -10 = 0
The solution reads as follows, but I do not understand it.
x squared + 3x = 10
4x squared + 12x = 40
4x squared + 12x +9 = 40 + 9
4x squared + 12x +9 = 49
2x + 3 = +-7
2x + 3 = 7
2x = 4
x = 2
and
2x + 3 = -7
2x = -10
x = -5
Is there anyway you can explain this method to me? I do not understand where the x squared turned into the 4x squared and where the 12x went later in the problem. Thank you in advance for your help.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
This so-called "Indian" method is just a variation of completing the square. I suspect you are being introduced to the method as a practical illustration of the fact that in mathematics, as well as most other fields of human endeavor, there is almost always more than one way to skin a cat.
In general, the Indian Method is as follows:
Given
Step 1: Add the opposite of the constant term to both sides of the equation.
Step 2: Multiply both sides by 4 times the original lead coefficient:
Step 3: Square the original coefficient on the  term and add that to both sides:
Step 4: The trinomial in the LHS is now a perfect square, so factor it:
^2\ =\ b^2\ -\ 4ac)
Step 5: Take the square root of both sides, remembering to consider both the positive and negative roots:
Step 6: Add the opposite of the constant term in the LHS to both sides and then divide both sides by the coefficient on the term in the LHS
Which you should recognize as the quadratic formula.
Given
Note that
 ,  , and 
Step 1: Add the opposite of the constant term to both sides of the equation.
Step 2: Multiply both sides by 4 times the original lead coefficient ( ):
x\ =\ 4\,\cdot\,10)
Step 3: Square the original coefficient on the term and add that to both sides ( ):
Step 4: The trinomial in the LHS is now a perfect square, so factor it:
^2\ =\ 49)
Step 5: Take the square root of both sides, remembering to consider both the positive and negative roots:
Step 6: Add the opposite of the constant term in the LHS to both sides and then divide both sides by the coefficient on the term in the LHS
or
Check your work:
If  and  are the roots of:
,
then
(x\ -\ \beta)\ =\ ax^2\ +\ bx\ +\ c)
So use FOIL to multiply
)(x\ -\ 2)\ =\ (x\ +\ 5)(x\ -\ 2)) to verify that the product is indeed
Verification left as an exercise for the student.
John
My calculator said it, I believe it, that settles it
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