Week_3_Assignment_Guidance (2)

# Week_3_Assignment_Guidance (2) - Indian Method and...

This preview shows pages 1–6. Sign up to view the full content.

Click to edit Master subtitle style 6/7/11 Indian Method and Quadratic Formula Math 126 Assignment 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6/7/11 Indian Method (Solving problem c: Step a) X2 + 12X – 64 Step a) Move constant term to right. The constant term is the term with no X variable beside it. The constant term is 64 X2 + 12X = 64
6/7/11 Indian Method (Solving problem c: Step b) Step b) Multiply each term in the equation by four times the coefficient of the X2 term. The X2 coefficient in this case is 1. If there is no number in front of the X2 term, then the number is 1. 1 X2 + 12X – 64. So four times 1 equals 4 ( 4 )X2 + (12 *4 )X = 64 *4 = 4X2 + 48X

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6/7/11 Indian Method (Solving problem c: Step c) Square the coefficient of the original x term and add it to both sides of the equation. Here is the original equation X2 + 12X – 64 The X term is 12, so 122 = 144. So add 144 to both sides. Start from where you left off at step b.
6/7/11 Indian Method (Solving problem c: Step d) Take square of both sides

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 06/07/2011 for the course EMPLOYEE L BUS372 taught by Professor Deniseantoon during the Spring '11 term at Ashford University.

### Page1 / 10

Week_3_Assignment_Guidance (2) - Indian Method and...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online