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OpenStudy (anonymous):

X^2 - 2x - 13 = 0 using the India method for solving quadratic equations

OpenStudy (anonymous):

Haven't heard that method before... could you describe it to us?

OpenStudy (anonymous):

The steps are (a) Move the constant term to the right side of the equation. (b) Multiply each term in the equation by four times the coefficient of the x2 term. (c) Square the coefficient of the original x term and add it to both sides of the equation. (d) Take the square root of both sides. (e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x. (f) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.

OpenStudy (anonymous):

This is what I have so far Step (a) Move the constant term to right of equation. x² - x2 – 13 = 0 – x² - x2 = 13 Step (b) Multiply each term in the equation by 4 times the coefficient of the x² term. (To do this mutiply both sides of the equation by 4 • 1 = 4 ) 4x² - 8x + 4 = 52 Step (c) Square the coefficient of the original x term and add it to both sides of the equation. The coefficient of the original term is 2 (2 squared is equal to 4, add it to both sides of your equation) 4x² - 8x + 4 = 56 Step (d) Take the square root of both sides. (2x–2)•(2x-2) = 4x² - 4x – 4x + 4 which equals the left side expression 4x² - 8x + 4 to then make the equation : (2x – 2)² = 56. Then take the square root of both sides to equal: 2x – 2 = +/- √56 divide both sides of the equation by 2 to get: x = √(56)+2)/2 Step (e): Set the left side of the equation to the positive square root of the number on the right side and solve for x you get: 2x - 2 = √56 add 2 to both sides of the equation to get: 2x = √56 + 2

OpenStudy (anonymous):

Okay, I'm unfamiliar with this method, but I think that for the final step you just set 2x-2 equal to positive root 56 (approximately 7.48) and solve, then do the same with negative 7.48. I'm getting x=4.74 and x=-2.74 Does that help or not what you were looking for?

OpenStudy (anonymous):

Let me work it into the quadratic formula to see if I get the same answer...

OpenStudy (anonymous):

possibly I will work it through I am trying to understand it myself still. ugh

OpenStudy (anonymous):

Yeah, same thing, but that's a really weird way of solving it :) So is your teacher making you learn this or this for fun?

OpenStudy (anonymous):

Teacher here is an example from the book Example: Solve x2 + 3x − 10 = 0. x2 + 3x = 10 4x2 + 12x = 40 4x2 + 12x + 9 = 40 + 9 4x2 + 12x + 9 = 49 2x + 3 = ±7 2x + 3 = 7 2x = 4 x = 2 2x + 3 = −7 2x = −10 x = −5

OpenStudy (anonymous):

But that's factorable. The one in your problem wasn't; that's why you got decimals

OpenStudy (anonymous):

I was following along with someone elses work suggestion and that is how they came up with the work. Now I am so confused

OpenStudy (anonymous):

Actually, believe it or not, I found the same problem online when I looked up India method. Here: http://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.251878.html

OpenStudy (anonymous):

wow, thank you I will look at that.

OpenStudy (anonymous):

disclamer: the guy who solved it said to use quadratic formula because it's not factorable. Sorry.

OpenStudy (anonymous):

Thats okay, I will check it out and see what it looks like and if it helps me understand it anymore.

OpenStudy (anonymous):

Thank you for your help :)

Directrix (directrix):

This technique appears to be solving quadratics by completing the square but without the fractions. X^2 + x + 1 is not factorable over the set of Reals. Solution by India Method: X^2 + x + 1 = 0 X^2 + x = -1 4x^2 + 4x = -4 4x^2 +4x + 1 = -4 + 1 (2x + 1)^2 = -3 2x + 1 = √ (-3) 2x = -1 - √ (-3) X = -1/2 – (i√ 3) / 2 or likewise, X = -1/2 + (i√ 3) / 2

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