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# Completing the square watch

1. Can you check what I’ve done posted below please. After that I also need to know the maximum value of 1/(2x^2+4x+9)
3. (Original post by Reece.W.J)
Two things:

1. You can check it for yourself by expanding it again. Useful skill for the exam.

2. For your follow up question, you posted something very similar here -

https://www.thestudentroom.co.uk/sho...2#post74126882

So what in particular are you finding difficult? Just the first part being correct?
4. (Original post by Kevin De Bruyne)
Two things:

1. You can check it for yourself by expanding it again. Useful skill for the exam.

2. For your follow up question, you posted something very similar here -

https://www.thestudentroom.co.uk/sho...2#post74126882

So what in particular are you finding difficult? Just the first part being correct?
So is the maximum value 1/(-6/4)
5. I've attempted the question myself but I managed to get a different answer:

1) 2[x^2 + 2x + 5/2] = 0
2) 2[(x^2 + 1)^2 - 1^2 + 5/2] = 0
3) 2[(x + 1)^2 + 3/2] = 0
4) (x + 1)^2 + 3 = 0

Therefore my turning points would be:
x = - 1
y = 3
or ( -1, 3 )

To calculate your turning points, your X value is the value inside the bracket and your Y value is the value outside.
I've also checked these using a calculator (casio fx-991EX) and got the same values I had worked out.

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Updated: October 15, 2017
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