studenthelp143 Posts: 7, Reputation: 1 New Member #1 Mar 28, 2011, 06:47 PM
How to solve quadratic equation x2-2x-3=0 using Indian Method?
Need detailed steps and solutions using the indian method.

Thanks
 jcaron2 Posts: 986, Reputation: 204 Senior Member #2 Mar 28, 2011, 08:01 PM
$x^2-2x-3=0$

First, move the constant term to the right side of the equation:

$x^2-2x=3$

Now multiply everything by four times the original coefficient of the $x^2$ term (i.e. 4 x 1 = 4):

$4x^2-8x=12$

Next, add the square of the original coefficient of x to both sides (i.e. $[-2]^2=4$):

$4x^2-8x+4=16$

This makes the left side of the equation a perfect square. So now you can take the square root of both sides:

$\sqrt{4x^2-8x+4}=\sqrt{16}$

$\sqrt{(2x-2)^2}=\sqrt{16}$

$2x-2=\pm 4$

$x=-1$ or $x=3$

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