Question

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

Answer 1

what is the 'India method'? sorry, I'm curious..

Answer 2

I googled it:
(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 x^2 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 to the positive square root of the number on the right side and solve for x
(f) Set the left side of the equation to the negative square root of the number on the right side of the equation and solve for x

Answer 3

I'm having trouble figuring out the last part of the problem

Answer 4

what do u have so far?

Answer 5

I am at step d I have 4x^2-2x+8=48

Answer 6

I am not sure how to get step e) and f)

Answer 7

hmm.. sorry, I'm not familiar with this method. maybe @dpaInc @Calcmathlete @eliassaab @Mertsj @jim_thompson5910 can help, they're smart (:

Answer 8

Ok. Thank you.

Answer 9

x^2 - 2x - 13 = 0


(a) Move the constant term to the right side of the equation


x^2 - 2x = 13


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(b) Multiply each term in the equation by four times the coefficient of the x^2 term


4x^2 - 8x = 52

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(c) Square the coefficient of the original x term and add it to both sides of the equation


4x^2 - 8x + 4 = 52 + 4


4x^2 - 8x + 4 = 56


(2x-2)^2 = 56


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(d) Take the square root of both sides


2x-2 = +-sqrt(56)


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(e) Set the left side of the equation to the positive square root of the number on the right side and solve for x



2x-2 = sqrt(56)


2x = 2 + sqrt(56)


x = (2 + sqrt(56))/2


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(f) Set the left side of the equation to the negative square root of the number on the right side of the equation and solve for x


2x-2 = -sqrt(56)


2x = 2 - sqrt(56)


x = (2 - sqrt(56))/2

Answer 10

\[
x^2 - 2x - 13 = 0 \\
x^2 -2 x + 1 -1 - 13=0
(x-1)^2 -14 =0 \\
(x-1)^2 = 14\\
x-1 =\pm \sqrt {14}\\
x = 1 \pm \sqrt {14}
\]

Answer 11

With each root, you can simplify further


x = (2 + sqrt(56))/2


x = (2 + sqrt(4*14))/2


x = (2 + sqrt(4)*sqrt(14))/2


x = (2 + 2*sqrt(14))/2


x = 1 + sqrt(14)

------------------------------


x = (2 - sqrt(56))/2


x = (2 - sqrt(4*14))/2


x = (2 - sqrt(4)*sqrt(14))/2


x = (2 - 2*sqrt(14))/2


x = 1 - sqrt(14)

Answer 12

\[
x^2 - 2x - 13 = 0 \\
x^2 -2 x + 1 -1 - 13=0\\
(x-1)^2 -14 =0 \\
(x-1)^2 = 14\\
x-1 =\pm \sqrt {14}\\
x = 1 \pm \sqrt {14}
\]
I forgot to put the \\ in the latex code in my post above.

Answer 13

in part f) of the equation though it says to 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.

Answer 14

that's exactly what's shown in the steps above

Answer 15

Ok. Sorry, I was just having a little trouble understanding all of the symbols

Answer 16

i see