The Student Room Group

completing the square

So ive completed the square of an equation to find:



but is the minimum is 2??
Reply 1
What was the original question?
Reply 2
Original post by razzor
What was the original question?


complete the square and find minimum of
x^4 +2x^2 +y^4 - 2y^2 +3
Reply 3
You are correct that the minimum is 2.
Reply 4
Original post by notnek
You are correct that the minimum is 2.


I don't see why?
Reply 5
Protip: don't link images straight from Wolfram, they clear the image out so it's broken on TSR after a few minutes.

Anyways, I presume you've got to (x2+1)2+(y21)2+1(x^2 + 1)^2 + (y^2 - 1)^2 + 1, now you want to make this expression as small as possible, since you're adding together three terms your best bet is to make each term as small as possible.

Let's try making the "11" as small as possible: well it's a constant. So it's stuck as 1.

Let's try making (y21)2y^2 -1)^2 as small as possible. Well it's essentially (something)2(something)^2 - so the smallest it can ever get is by making something=0 (i.e: y=1y = 1) to get (121)2=02=0(1^2-1)^2 = 0^2 = 0.

It won't get any smaller than that. (something)^2 is always positive, your only hope to make it small is by making it 0.

Now we'll want the same tactic to work with (x2+1)2(x^2 + 1)^2. Here, since x2x^2 is a squared term, the smallest you can make it is 0. So the inside becomes (0^2 + 1)^2 = 1^2 = 1. That's the smallest you can make it.

So all in all, the smallest your entire expression can get is 1 + 0 + 1 = 2.
Completing the square, x4+2x2+y42y2+3=(x2+1)2+(y21)2+1x^4+2x^2+y^4-2y^2+3=(x^2+1)^2+(y^2-1)^2+1.

The minimum of (x2+1)2(x^2+1)^2 is 11 (when x=0x=0) and the minimum of (y21)2(y^2-1)^2 is 0 0 (when y=1y=1).

Therefore the minimum value of the required expression is 1+0+1=21+0+1=2.
Original post by Zacken
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Yeah this is more detailed; it's a good insight into how to work out problems without memorising random techniques that you could easily forget.
Reply 8
Original post by IrrationalRoot
Yeah this is more detailed; it's a good insight into how to work out problems without memorising random techniques that you could easily forget.


Goes to show how the A-Level system is broken. A student would be able to deal with (x+a)^2 + k just fine because that's what they've been taught, but can't apply the exact same reasoning to (x^2 + a)^2 + (y^2 + b)^2 + k.
Reply 9
Sorry, when I got to that stage I wrote + instead of - inside the y bracket, so I could've worked it out all along! haha
Original post by Zacken
Goes to show how the A-Level system is broken. A student would be able to deal with (x+a)^2 + k just fine because that's what they've been taught, but can't apply the exact same reasoning to (x^2 + a)^2 + (y^2 + b)^2 + k.


Just a silly mistake in transcription! It was an easy question, but I got '3' instead of '2' from the error, so wanted to check!
Reply 11
Wasn't talking about you with that post - don't worry. :tongue:
Reply 12
Original post by Zacken
Goes to show how the A-Level system is broken. A student would be able to deal with (x+a)^2 + k just fine because that's what they've been taught, but can't apply the exact same reasoning to (x^2 + a)^2 + (y^2 + b)^2 + k.

It's now been moved to GCSE:

"Deduce turning points by completing the square".

I'm sure teachers will continue to teach that the minimum of (x+a)2+b(x+a)^2 + b is (a,b)(-a, b) without further explanation.

But there is hope : I saw a specimen GCSE follow-up question that asked you to find the two turning points of (x24)2+2(x^2-4)^2 + 2.
Reply 13
Original post by notnek
It's now been moved to GCSE: [...]

But there is hope : I saw a specimen GCSE follow-up question that asked you to find the two turning points of (x24)2+2(x^2-4)^2 + 2.


That's good! It was in my AddMaths GCSE. Appreciating the reform for maths so far - hope it doesn't put off many students.
Original post by Zacken
Goes to show how the A-Level system is broken. A student would be able to deal with (x+a)^2 + k just fine because that's what they've been taught, but can't apply the exact same reasoning to (x^2 + a)^2 + (y^2 + b)^2 + k.


Yep teachers always teach all the content with no explanation. It's terrible, really.
And most students don't seem to bother to go out of their way to understand what they're learning; instead they just churn through the same questions with their memorised algorithms and then complain that A-Level maths is 'so hard'.
Original post by Zacken
Protip: don't link images straight from Wolfram, they clear the image out so it's broken on TSR after a few minutes.

Anyways, I presume you've got to (x2+1)2+(y21)2+1(x^2 + 1)^2 + (y^2 - 1)^2 + 1, now you want to make this expression as small as possible, since you're adding together three terms your best bet is to make each term as small as possible.

Let's try making the "11" as small as possible: well it's a constant. So it's stuck as 1.

Let's try making (y21)2y^2 -1)^2 as small as possible. Well it's essentially (something)2(something)^2 - so the smallest it can ever get is by making something=0 (i.e: y=1y = 1) to get (121)2=02=0(1^2-1)^2 = 0^2 = 0.

It won't get any smaller than that. (something)^2 is always positive, your only hope to make it small is by making it 0.

Now we'll want the same tactic to work with (x2+1)2(x^2 + 1)^2. Here, since x2x^2 is a squared term, the smallest you can make it is 0. So the inside becomes (0^2 + 1)^2 = 1^2 = 1. That's the smallest you can make it.

So all in all, the smallest your entire expression can get is 1 + 0 + 1 = 2.

brilliant explanation, ive always known what to do but didn't really know why I had to do it (I sort of knew but that well), but this is brilliant and makes even more sense
Reply 16
Original post by metrize
brilliant explanation, ive always known what to do but didn't really know why I had to do it (I sort of knew but that well), but this is brilliant and makes even more sense


Thanks

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