Download Steps for the Indian method

YOUR NAME HERE
DATE
ANSWERS MUST BE HIGHLIGHTED IN YELLOW
Hint: please see additional guidance for using the Indian
method at the bottom of this document.
Intro Paragraph:
(TYPE YOUR INTRO PARAGRAPH HERE)
Project # 1
Task: Solve quadratic equations using the Indian Method. For Project #1, work only equations
(a) and (c), but complete all 6 steps (a-f) as shown in the example in the textbook. Please note
you must show your work and calculations.
Steps for the Indian method:
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 for 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.
Equation a:
x^2 – 2x – 13 = 0
Step a Work: X^2-2x=13
Step b Work: 4*x^2-2*4x=13*4 4x^2-8x=52
Step c Work: 4x^2-8x+4=52+4 4x^2-8x+4=56
Step d Work: 2𝑥 − 2 = ≠ √56
Step e Work: 2x-2=sqrt (56) 2x-2=2sqrt (14) x-1=sqrt (14) x=sqrt (14) +1
Step f Work: 2x-2=-sqrt (56) 2x-2=-2sqrt (14) x-1=-sqrt (14) x=-sqrt (14) +1
Solution (highlight in yellow): x = sqrt (14) +1 and x = x=-sqrt (14) +1
Equation c:
x^2 +12x – 64 = 0
Step a Work: x^2 +12x =64
Step b Work:
Step c Work:
Step d Work:
Step e Work:
Step f Work:
Solution (highlight in yellow): x = and x =
Project # 2
Task: To produce prime numbers as solutions by substituting x in the formula
2
x – x + 41
Please note you MUST show your work (plugging the number into the formula and
simplifying).
Here is an example of showing your work. Let’s say I start with 3 as one of my odd
numbers of choice. I need to plug 3 into the formula like this:
32 – 3 + 41
= 9 – 3 + 41
= 6 + 41
= 47
Is 47 prime? Yes! In this case, the formula produced a prime number!
Please perform the above steps for all five of your numbers (0, two even numbers of your
choice, and two odd numbers of your choice).
*** Note: this is NOT an equation. Instead, it is a FORMULA. Please do NOT set the
formula equal to 0 and try to factor, because this is not correct. 
Step 1: First even number of your choice: plug in number into formula and simplify:
ANSWER HERE
Step 2: Second even number of your choice: plug in number into formula and simplify:
ANSWER HERE
Step 3: First odd number of your choice: plug in number into formula and simplify:
ANSWER HERE
Step 4: Second odd number of your choice: plug in number into formula and simplify:
ANSWER HERE
Step 5: Plug in 0 into formula and simplify:
ANSWER HERE
Conclusion Paragraph:
(TYPE YOUR CONCLUSION PARAGRAPH HERE)
Please include any references as necessary (in APA format) on the last page (Works Cited).
STUCK ON THE INDIAN METHOD???
TRY THIS!!
Here is an example that illustrates how to tackle the hardest part of the Indian method – taking the
square root of both sides!
Let’s say we start with the following equation:
4x^2 + 12x + 9 = 49
One of the methods to solve quadratic equations is taking the square root of both sides. By rule,
when you take the square root, it is always plus or minus because for example both (-2)^2 = 4
and (+2)^2 = 4. Because both can be 4 (whether the 2 was positive or negative) is the reason we
have the plus or minus (written as +/-).
4x^2 + 12x + 9 = 49
To take the square root of both sides, we need one side to be a perfect square - so factor first.
(2x + 3)(2x +3) = 4x^2 + 12x + 9
So, 4x^2 + 12x + 9 = 49 can be written as
(2x +3)^2 = 49
Now, take the square root of both sides
sqrt[(2x +3)^2 ]= +/- sqrt(49)
(2x + 3) = +/- 7
2x + 3 = 7 and
2x + 3 = -7
As you read in the text - solving quadratics always gives 2 results.
Let's solve both for the values of x:
2x + 3 = 7 and
2x + 3 = -7
From the first, x = 2
From the second, x = -5
Let's check both solutions:
Does x = 2 work? Put it into the original equation to see:
4x^2 + 12x + 9 = 49
4(2)^2 + 12(2) + 9 = 49
4(4) + 24 + 9 = 49?
16 + 24 + 9 =49? Yes!
Let's check x = -5
4x^2 + 12x + 9 = 49
4(-5)^2 + 12(-5) + 9 = 49
4(25) - 60 + 9 = 49?
100 - 60 + 9
49 = 49 Yeah!
NOTE: You can also ALWAYS solve quadratic equations (equations with x^2 terms) using the
quadratic formula or with factoring. You can check your solution this way:
4x^2 + 12x + 9 = 49
Subtract 49 from both sides to get the "0" on the right.
4x^2 + 12x + 9- 49 = 0
4x^2 + 12x -40 = 0
4(x^2 +3x - 10) = 0
4(x -2)(x+ 5) = 0
x = 2, -5 (same result :)
Tutorial: http://www.purplemath.com/modules/solvquad2.htm
2nd tutorial: http://www.purplemath.com/modules/sqrquad.htm