April 8th, 2017, 06:42 AM  #1 
Member Joined: Jan 2015 From: London Posts: 98 Thanks: 2  Completing the Square
$\displaystyle x^2+6x2$ 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 
Member Joined: Jan 2015 From: London Posts: 98 Thanks: 2  
April 8th, 2017, 07:10 AM  #4 
Senior Member Joined: May 2016 From: USA Posts: 574 Thanks: 248  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 
Math Team Joined: Jul 2011 From: Texas Posts: 2,422 Thanks: 1189  
April 8th, 2017, 07:24 AM  #6 
Member Joined: Jan 2015 From: London Posts: 98 Thanks: 2  
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. $\displaystyle 2x^2+20x+3$ $\displaystyle 2(x^2+10)+3$ $\displaystyle 2[(x+5)^225]+3$ $\displaystyle 2(x+5)^250+3$ $\displaystyle 2(x+5)^247$ Then: $\displaystyle 2(x+5)^247=0$ $\displaystyle 2(x+5)^2=47$ $\displaystyle (x+5)^2=23.5$ $\displaystyle (x+5)=\pm\sqrt{23.5}$ $\displaystyle x=5+\sqrt{23.5}$ 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 $\displaystyle x^2$ $\displaystyle 26xx^2$ goes to: $\displaystyle x^26x+2$ so: $\displaystyle (x3)^2$ because $\displaystyle x^2$ goes to $\displaystyle x$ when being timed together? $\displaystyle (x3)^29+2$ because your taking away the +9 when you do $\displaystyle (x3)^2$ to $\displaystyle x^26x+9$ So I end up with this: $\displaystyle (x3)^27$ 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 
$x^26x+2$ $(x^2+6x2)$ $\left[(x^2+6x+9)  9  2\right]$ $\left[(x+3)^2  11\right]$ $(x+3)^2 + 11$ 

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