Solving Quadratic Equations That Came From India.

¶ … solving quadratic equations that came from India. The traditional method taught in American schools uses factoring. The Indian method isolated the constant on the right side of the equation and solves for X without using parentheses or cross multiplication, instead using squares and square roots (Bluman, 2011, p. 397). The steps 1 -- 6 for solving each equation are taken from Mathematics in Our World by A.G. Bluman.

a) x2 -- 2x -- 13 = 0

Move the constant term to the right side of the equation.

Multiply each term in the equation by four times the coefficient of the x squared term.

Square the coefficient of the original x term and add it to both sides of the equation.

4x2 -- 8x + 4 = 56

Take the square root of both sides.

2)2 = (22*2*7)

Divide each side by 2 simplify equation: x-1 = (1) * (14)

Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.

x - 1 = +Sqrt (14)

x = 1 + Sqrt (14)

x = 1 + 3.74166

x = 4.74166

6. 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.

x - 1 = -Sqrt (14)

x - 1+ 1 = -Sqrt (14)

x = 1 -- 3.74166

x = -2.74166

b) 4x2 -- 4x + 3 = 0

1. Move the constant term to the right side of the equation.

4x2 -- 4x = -3

2. Multiply each term in the equation by four times the coefficient of the x squared term.

4 * 4 = 16

16(4x2) -- 16(4x) = -3 * 16

64x2 -- 64x + -48

3. Square the coefficient of the original x term and add it to both sides of the equation.

64x2 -- 64x + 16 = -48 + 16

64x2 -- 64x + 16 = -32

4. Take the square root of both sides.

(8x -- 8)2 = Sqrt (32)

Divide each side by 8 to simplify the equation: x -- 1 = 4

5. Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.

x -- 1 = +Sqrt (4)

x = 2 + 1

6. 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.

x -- 1 = -Sqrt (4)

x = -2 + 1

x = -1

c) x2 + 12 -- 64 = 0

1. Move the constant term to the right side of the equation.

x2 + 12x -- 64 = 0

2. Multiply each term in the equation by four times the coefficient of the x squared term.

4x2 + 48x = 256

3. Square the coefficient of the original x term and add it to both sides of the equation.

4x2 + 48x + 144 = 256 + 144

4. Take the square root of both sides.

(2x +12)2 = (202) Divide each side by 2 to simplify: x + 6 = 10

5. Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.

x + 6 = 10

x = 10 -- 6 OR x = 4

6. 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.

x + 6 = -10

x = -10 -- 6 OR x = -16

d) 2x2 -- 3x -- 5 = 0

1. Move the constant term to the right side of the equation.

2x2 -- 3x = 5

2. Multiply each term in the equation by four times the coefficient of the x squared term.

8(2x2) -- 8(3x) = 5 * 8

16x2 -- 24x = 40

3. Square the coefficient of the original x term and add it to both sides of the equation.

16x2 -- 24x + 9 = 49

4. Take the square root of both sides.

(4x -- 3)2 = Sqrt (49)

5. Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.

4x -- 3 = 7

4x = 10

x = .4

6. 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.

4x -- 3 = -7

4x = -4

x = -1

Project 2 tests a formula that yields prime numbers. For this assignment, 5 numbers are selected:

zero, two even numbers ( 4 and 6) and two odd numbers (5 and 7).

The formula: x2 -- x + 41 (Bluman, 2011, p. 397).

x2 -- x + 41 where x = 0