1. ## 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: $\displaystyle x^2-6x-1=0$ Then I know I would want to switch it to:$\displaystyle x^2-6x=1$ 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

2. 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: $\displaystyle x^2-6x-1=0$ Then I know I would want to switch it to:$\displaystyle x^2-6x=1$ 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.

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

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

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

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

simplify

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

the square has been completed

3. 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.

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

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

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

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

simplify

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

the square has been completed

ok thanks!

so how would i find out what x equals?

4. Originally Posted by deathtolife04
ok thanks!

so how would i find out what x equals?
solve for x the old-fashion way. try it

5. Originally Posted by Jhevon
solve for x the old-fashion way. try it
what would I do after I get to $\displaystyle x^2-6x=1$?

6. Originally Posted by deathtolife04
what would I do after I get to $\displaystyle x^2-6x=1$?
why are you going back there? did you let me complete the square for nothing? start from $\displaystyle (x - 3)^2 - 10 = 0$

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

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

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 $\displaystyle (x-3)^2=10$

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

Could it just be x=the square root of 10+3?

8. 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.

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

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

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

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

simplify

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

the square has been completed
Originally Posted by deathtolife04
ok thanks!

so how would i find out what x equals?
To continue:
$\displaystyle (x - 3)^2 - 10 = 0$

$\displaystyle (x - 3)^2 = 10$

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

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

-Dan