Question 148733: I have to solve the equation using the Indian method which is: a) Move the constant to the right of the equation, b) multiply each term by 4 times the coefficient of the x^ term, c) Square the coefficient of the x term and put it on 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 sqrt of the number on the right side - solve for x, and f)set the left side of the equation equal to the negative sqrt of the number on the right side - solve for x. I am stumped!!!
The equation is: x^ + 12x - 64 = 0
Please help, Ellen
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I have to solve the equation (x^ + 12x - 64 = 0) using the Indian method which is:
a) Move the constant to the right of the equation,
x^2 +12x = 64
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b) multiply each term by 4 times the coefficient of the x^2 term,
4x^2 + 48x = 4*64
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c) Square the coefficient of the x term and put it on both sides of the equation,
4x^2 + (48)^2 = 4*64 + 48^2
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d) take the square root of both sides,
Comment: That would be a mess; I think you have your instructions wrong.
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e) set the left side of the equation equal to the positive sqrt of the number on the right side - solve for x, and
f) set the left side of the equation equal to the negative sqrt of the number on the right side - solve for x. I am stumped!!!
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Let's try again:
x^2 + 12x - 64 = 0
x^2 + 12x = 64
Complete the square on the x-terms:
x^2 + 12x + 6^2 = 64 + 6^2
Factor to get:
(x+6)^2 = 100
Take the square root:
x+6 = 10 or x+6 = -10
x = 4 or x = -16
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Cheers,
Stan H.
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