Download An interesting method for solving quadratic equations

An interesting method for solving quadratic equations came from India. The steps are
(a) Move the constant terms 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 equal to the negative square root of the number on the
right side of the equation and solve for x.
Solution:
1. x^2 -2x - 13 = 0
Move the constant term to the right side of the equation
x^2 - 2x = 13
Multiply each side by 4*1 = 4.
4x^2 - 8x = 52
Add (-2)^2 = 4 to each side.
4x^2 - 8x + 4 = 52 + 4
(2x - 2)^2 = 56
2x - 2 = ±√56
2x - 2 = ±2√14
When 2x - 2 = 2√14
2x = 2 + 2√14
x = 1 + √14
When 2x - 2 = - 2√14
2x = 2 - 2√14
x = 1 - √14
Answer: x = 1 - √14, 1 + √14
2. 4x^2 - 4x + 3 = 0
Move the constant term to the right side of the equation
4x^2 - 4x = - 3
Multiply by 4*4 = 16 on each side
64x^2 - 64x = -48
Add (-4)^2 = 16 to each side
64x^2 - 64x + 16 = -48 + 16
(8x - 4)^2 = -32
8x - 4 = ±√-32
8x - 4 = ±4i√2
When 8x - 4 = 4i√2
8x = 4 + 4i√2
x = (1/2) + i(√2/2)
When 8x - 4 = -4i√2
8x = 4 - 4i√2
x = (1/2) - i(√2/2)
Answer: x = (1/2) + i(√2/2), (1/2) - i(√2/2)
3. x^2 + 12x - 64 = 0
Move the constant term to the right side of the equation
x^2 + 12x = 64
Multiply by 4*1 = 4 on each side
4x^2 + 48x = 256
Add 12^2 =144 to each side
4x^2 + 48x + 144 = 256+144
(2x + 12)^2 = 400
2x + 12 = ±20
When 2x + 12 = 20
2x = 20 - 12 = 8
x=4
When 2x + 12 = -20
2x = -20 - 12 = -32
x = -16
Answer: x = -16, 4
4. 2x^2 - 3x -5 = 0
Move the constant term to the right side of the equation
2x^2 - 3x = 5
Multiply by 2*4 = 8 on each side
16x^2 - 24x = 40
Add (-3)^2 = 9 to each side
16x^2 - 24x + 9 = 40 + 9
(4x - 3)^2 = 49
4x - 3 = ±7
When 4x - 3 = 7
4x = 10
x = 10/4 = 5/2
When 4x - 3 = -7
4x = -4
x = -1
Answer: x = -1, 5/2
Mathematicians have been searching for a formula that yields prime numbers. One
such formula was x^2 - x +41. Select some numbers for x, substitute them in the
formula, and see if prime numbers occur. Try to find a number for x that when
substituted in the formula yields a composite number.
Solution:
Let us take two even numbers 2 and 4 and two odd numbers 7 and 9.
Let P(x) = x^2 - x +41
P(2) = 2^2 – 2 + 41 = 4 – 2 + 41 = 43, which is a prime number.
P(4) = 4^2 – 4 + 41 = 16 – 4 + 41 = 53, which is a prime number.
P(0) = 0^2 – 0 + 41 = 41, which is a prime number.
P(7) = 7^2 – 7 + 41 = 49 – 7 + 41 = 83, which is a prime number.
P(9) = 9^2 – 9 + 41 = 81 – 9 + 41 = 113, which is a prime number.
We can note that in all the cases prime numbers occur.
Let us try x = 41
P(41)
= 41^2 – 41 + 41
= 1681 – 41 + 41
= 1681, which is a composite number as 1681 = 41*41.