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 April 8th, 2017, 06:42 AM #1 Member   Joined: Jan 2015 From: London Posts: 98 Thanks: 2 Completing the Square $\displaystyle x^2+6x-2$ in the form $\displaystyle (x+p)^2+q$ That's the question and I know its fairly simple but want some help with the whole completing the square topic in general. Here's my working: Take $\displaystyle x^2+6$ and make it $\displaystyle (x+3)^2$ and then you have $\displaystyle -2$ at the end. Then you do: $\displaystyle (x+3)(x+3)$ to get $\displaystyle x^2+6x+9$ After you do something to the $\displaystyle 9$ but I forgot? Any help on the topic for aqa further maths igcse would be helpful! Thanks
 April 8th, 2017, 06:55 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,422 Thanks: 1189 $x^2+ 6x - 2$ $(x^2 + 6x + 9) - 9 - 2$ $(x+3)^2 - 11$
April 8th, 2017, 06:56 AM   #3
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Quote:
 Originally Posted by skeeter $x^2+ 6x - 2$ $(x^2 + 6x + 9) - 9 - 2$ $(x+3)^2 - 11$
Why do you subtract the 9?

April 8th, 2017, 07:10 AM   #4
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Quote:
 Originally Posted by jamesbrown Why do you subtract the 9?
Note that you are dealing with an expression, not an equation.

You can change the form of an expression so long as you do not change its value.

$x^2 + 6x - 2 \equiv x^2 + 2(3)x - 2 \equiv x^2 + 2(3)x + 0 - 2 \equiv$

$x^2 + 2(3)x + (3^2 - 3^2) - 2 \equiv (x^2 + 2(3)x + 3^2) - 9 - 2 \equiv (x + 3)^2 - 11.$

April 8th, 2017, 07:15 AM   #5
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 Originally Posted by jamesbrown Why do you subtract the 9?
because I added 9 to complete the square ...

+9-9 = 0 ... adding 0 to an expression leaves it the same value.

April 8th, 2017, 07:24 AM   #6
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Quote:
 Originally Posted by skeeter because I added 9 to complete the square ... +9-9 = 0 ... adding 0 to an expression leaves it the same value.
That makes sense thanks a bunch!

 April 8th, 2017, 08:07 AM #7 Member   Joined: Jan 2015 From: London Posts: 98 Thanks: 2 Thanks for all your help! I can now do it - I think. Here's another question I did. $\displaystyle 2x^2+20x+3$ $\displaystyle 2(x^2+10)+3$ $\displaystyle 2[(x+5)^2-25]+3$ $\displaystyle 2(x+5)^2-50+3$ $\displaystyle 2(x+5)^2-47$ Then: $\displaystyle 2(x+5)^2-47=0$ $\displaystyle 2(x+5)^2=47$ $\displaystyle (x+5)^2=23.5$ $\displaystyle (x+5)=\pm\sqrt{23.5}$ $\displaystyle x=-5+\sqrt{23.5}$ or $\displaystyle x=-5-\sqrt{23.5}$ Last edited by skipjack; April 8th, 2017 at 08:20 AM.
 April 8th, 2017, 08:18 AM #8 Global Moderator   Joined: Dec 2006 Posts: 16,773 Thanks: 1234 In your second line of working, you should have $10x$ instead of just $10$.
 April 8th, 2017, 10:18 AM #9 Member   Joined: Jan 2015 From: London Posts: 98 Thanks: 2 I now have this other equation that I want to simplify to $\displaystyle (x+p)^2+q$ but it has a negative $\displaystyle x^2$ $\displaystyle 2-6x-x^2$ goes to: $\displaystyle -x^2-6x+2$ so: $\displaystyle (x-3)^2$ because $\displaystyle -x^2$ goes to $\displaystyle x$ when being timed together? $\displaystyle (x-3)^2-9+2$ because your taking away the +9 when you do $\displaystyle (x-3)^2$ to $\displaystyle x^2-6x+9$ So I end up with this: $\displaystyle (x-3)^2-7$ But I think I did it wrong? Can someone explain?
 April 8th, 2017, 11:10 AM #10 Math Team   Joined: Jul 2011 From: Texas Posts: 2,422 Thanks: 1189 $-x^2-6x+2$ $-(x^2+6x-2)$ $-\left[(x^2+6x+9) - 9 - 2\right]$ $-\left[(x+3)^2 - 11\right]$ $-(x+3)^2 + 11$

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