Completing the Square

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krispy

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Sep 12, 2006
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I'm a little sketchy on how to go about finishing this problem:

2x^2 + 9x - 1 = 0

Do I move the -1 over and try to solve from there? Or do I divide by 2 and solve?
 
krispy said:
I'm a little sketchy on how to go about finishing this problem:

2x^2 + 9x - 1 = 0

Do I move the -1 over and try to solve from there? Or do I divide by 2 and solve?
Click to expand...
Completing the square.

Given x^2 + 10x - 4 = 0.
Transfer the -4 to the other side yielding x^2 + 10x = 4.
Add the square of 1/2 the coefficient of x to both sides yielding x^2 + 10x + 25 = 29.
Then, (x + 5)^2 = 29 or x + 5 = +/-sqrt29 or x = -5 +/- sqrt29.

Using the quadratic equation
x^2 + 10x - 4 = 0.
Therefore, x = [-10+/-sqrt100 + =16)]/2 or
= [-10+/-2sqrt29)]/2 or x = -5 +/-sqrt29.

Given 4x^2 + 4x -11
.........4x^2 + 4x = 11
Divide through by the coefficient of x^2
..........x^2 + x = 11/4
Add the square of 1/2 the coefficient of x to both sides
..........x^2 + x + 1/4 = 11/4 + 1/4 = 12/4 = 3
..........(x + 1/2)^2 = 3
...........x + 1/2 = +/-3
...........x = -1/2 +/-3
 
Alright, so when I follow your example for completing the square I get:

(x + 3)(2x + 3) = 10

...Is that the right answer? It seems like it's in the wrong form.
 
Your answer is entirely useless. If you are to get something useful, that multiplication mustbe zero (0). "10" simply will not do.

x+3
2x+3

What have these two DIFFERENT terms to do with a SQUARE?

The is the fundamental principle: (a+b)^2 = a^2 + 2*a*b + b^2
 
You're right, my answer is entirely useless. So, instead of moving the -1 over I should divide by 2, right? Then what?
 
2x^2 + 9x - 1 = 0

Divide by 2

x^2 + (9/2)x - (1/2) = 0

Add 1/2

x^2 + (9/2)x = 1/2

Calculate the magic term you need for a perfect square.

(9/2)/2 = 9/4
(9/4)^2 = 81/16

Add 81/16

x^2 + (9/2)x + (81/16) = (1/2) + (81/16)

Simplify

[x + (9/4)]^2 = (89/16)

Can you finish?
 
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