Question 373661
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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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


Step 1:  Add the opposite of the constant term to both sides of the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ =\ -c]


Step 2:  Multiply both sides by 4 times the original lead coefficient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a^2x^2\ +\ 4abx\ =\ -4ac]


Step 3:  Square the original coefficient on the *[tex \Large x] term and add that to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a^2x^2\ +\ 4abx\ +\ b^2\ =\ b^2\ -\ 4ac]


Step 4:  The trinomial in the LHS is now a perfect square, so factor it:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(2ax\ +\ b\right)^2\ =\ b^2\ -\ 4ac]


Step 5:  Take the square root of both sides, remembering to consider both the positive and negative roots:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2ax\ +\ b\ =\ \pm\sqrt{b^2\ -\ 4ac}]


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 *[tex \Large x] term in the LHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


Which you should recognize as the quadratic formula.


Given


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 3x\ -\ 10\ =\ 0]


Note that 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ 3\ \ ], and *[tex \LARGE c\ =\ -10]


Step 1:  Add the opposite of the constant term to both sides of the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 3x\ =\ 10]


Step 2:  Multiply both sides by 4 times the original lead coefficient (*[tex \Large 4\,\cdot\,1\ =\ 4]):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ +\ (4\,\cdot\,3)x\ =\ 4\,\cdot\,10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ +\ 12x\ =\ 40]



Step 3:  Square the original coefficient on the *[tex \Large x] term and add that to both sides (*[tex \Large b\ =\ 3\ \rightarrow\ b^2\ =\ 9]):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ +\ 12x\ +\ 9\ =\ 9\ +\ 40]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ +\ 12x\ +\ 9\ =\ 49]


Step 4:  The trinomial in the LHS is now a perfect square, so factor it:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(2x\ +\ 3\right)^2\ =\ 49]


Step 5:  Take the square root of both sides, remembering to consider both the positive and negative roots:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 3\ =\ \pm\sqrt{49}]


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 *[tex \Large x] term in the LHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-3\ \pm\ \sqrt{49}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-10}{2}\ =\ -5]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{4}{2}\ =\ 2]


Check your work:


If *[tex \Large \alpha] and *[tex \Large \beta] are the roots of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0],


then 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ \alpha)(x\ -\ \beta)\ =\ ax^2\ +\ bx\ +\ c]


So use FOIL to multiply


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ (-5))(x\ -\ 2)\ =\ (x\ +\ 5)(x\ -\ 2)] to verify that the product is indeed


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 3x\ -\ 10\ =\ 0]


Verification left as an exercise for the student.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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