Completing the Square

• Aug 31st 2007, 09:30 PM
deathtolife04
Completing the Square
Can someone help me out and go through the steps for how to complete the square for this problem? I can never quite figure it out.

The problem is: $x^2-6x-1=0$ Then I know I would want to switch it to: $x^2-6x=1$ Normally I know I would divide everything by what is in front of the x, but in that case it's nothing (1), and dividing by 1 obviously wouldn't do anything. So what else do I need to do?

thanks
• Aug 31st 2007, 09:53 PM
Jhevon
Quote:

Originally Posted by deathtolife04
Can someone help me out and go through the steps for how to complete the square for this problem? I can never quite figure it out.

The problem is: $x^2-6x-1=0$ Then I know I would want to switch it to: $x^2-6x=1$ Normally I know I would divide everything by what is in front of the x, but in that case it's nothing (1), and dividing by 1 obviously wouldn't do anything. So what else do I need to do?

thanks

add and subtract the square of (1/2) the coefficient of x (that way, we're adding zero and hence not changing anything.

$x^2 - 6x - 1 = 0$

$\Rightarrow x^2 - 6x ~{ \color {red} + (-3)^2 - (-3)^2} - 1 = 0$

now, contract the first three terms as you have been taught

$\Rightarrow (x - 3)^2 -(-3)^2 - 1 = 0$

simplify

$\Rightarrow (x - 3)^2 - 10 = 0$

the square has been completed
• Aug 31st 2007, 10:17 PM
deathtolife04
Quote:

Originally Posted by Jhevon
add and subtract the square of (1/2) the coefficient of x (that way, we're adding zero and hence not changing anything.

$x^2 - 6x - 1 = 0$

$\Rightarrow x^2 - 6x ~{ \color {red} + (-3)^2 - (-3)^2} - 1 = 0$

now, contract the first three terms as you have been taught

$\Rightarrow (x - 3)^2 -(-3)^2 - 1 = 0$

simplify

$\Rightarrow (x - 3)^2 - 10 = 0$

the square has been completed

ok thanks!

so how would i find out what x equals?
• Aug 31st 2007, 10:18 PM
Jhevon
Quote:

Originally Posted by deathtolife04
ok thanks!

so how would i find out what x equals?

solve for x the old-fashion way. try it
• Aug 31st 2007, 10:24 PM
deathtolife04
Quote:

Originally Posted by Jhevon
solve for x the old-fashion way. try it

what would I do after I get to $x^2-6x=1$?
• Aug 31st 2007, 10:28 PM
Jhevon
Quote:

Originally Posted by deathtolife04
what would I do after I get to $x^2-6x=1$?

why are you going back there? did you let me complete the square for nothing? start from $(x - 3)^2 - 10 = 0$

remember, when we are solving for x, it means we want to get x by itself on one side of the equation, so do it
• Aug 31st 2007, 10:34 PM
deathtolife04
Quote:

Originally Posted by Jhevon
why are you going back there? did you let me complete the square for nothing? start from $(x - 3)^2 - 10 = 0$

remember, when we are solving for x, it means we want to get x by itself on one side of the equation, so do it

ok so $(x-3)^2=10$

where would i go from there to get x by itself? the () and $^2$ confuse me.

Could it just be x=the square root of 10+3?
• Aug 31st 2007, 10:36 PM
topsquark
Quote:

Originally Posted by Jhevon
add and subtract the square of (1/2) the coefficient of x (that way, we're adding zero and hence not changing anything.

$x^2 - 6x - 1 = 0$

$\Rightarrow x^2 - 6x ~{ \color {red} + (-3)^2 - (-3)^2} - 1 = 0$

now, contract the first three terms as you have been taught

$\Rightarrow (x - 3)^2 -(-3)^2 - 1 = 0$

simplify

$\Rightarrow (x - 3)^2 - 10 = 0$

the square has been completed

Quote:

Originally Posted by deathtolife04
ok thanks!

so how would i find out what x equals?

To continue:
$(x - 3)^2 - 10 = 0$

$(x - 3)^2 = 10$

$x - 3 = \pm \sqrt{10}$

$x = 3 \pm \sqrt{10}$

-Dan