Completing the Square

**deathtolife04** · August 31st 2007, 09:30 PM

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

**Jhevon** · August 31st 2007, 09:53 PM

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

**deathtolife04** · August 31st 2007, 10:17 PM

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?

**Jhevon** · August 31st 2007, 10:18 PM

Originally Posted by deathtolife04

ok thanks!

so how would i find out what x equals?

solve for x the old-fashion way. try it

**deathtolife04** · August 31st 2007, 10:24 PM

Originally Posted by Jhevon

solve for x the old-fashion way. try it

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

**Jhevon** · August 31st 2007, 10:28 PM

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

**deathtolife04** · August 31st 2007, 10:34 PM

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?

**topsquark** · August 31st 2007, 10:36 PM

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

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

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