Question 348179: Solving a quadratic equation with the Indian method:
Solve x^2+3x-10=0
x^2+3x=10
Next step is to multiply each term by four times the original coefficient of the x^2 term
4x^2+12x=40
Next square the original coefficient of x and add it to both sides of the equation:
4x^2+12x+9=40+9
4x^2+12x+9=49
Next is to take the square root of both sides, this is where I'm confused by my textbook:
I know that the square root of 4x^2 = 2x but the square root of 12x is not going to be a 'clean' number and the square root of 9 and 49 are 3 and 7... the textbook says that 4x^2+12x+9=49 taken to the square root is
2x + 3= +-7
Where did the 12x go?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Solve x^2+3x-10=0
x^2+3x=10
Next step is to multiply each term by four times the original coefficient of the x^2 term
4x^2+12x=40
Next square the original coefficient of x and add it to both sides of the equation:
4x^2+12x+9=40+9
4x^2+12x+9=49
Next is to take the square root of both sides
-------------------
The left side factors as (2x+3)^2
----
(2x+3)^2 = 49
Taking the square root of both sides you get:
2x+3 = 7 or 2x+3 = -7
2x = 4 or 2x = -10
x = 2 or x = -5
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Cheers,
Stan H.
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