This is a question on an AS core 1 paper. Im not really sure what its asking me to do.

Find constants a and b, such that for all values of x:

x^2 + 4x + 14 = (x + a)^2 +b

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- Jan 1st 2008, 11:16 PMnugiboyNeed help. Sort of Urgent!
This is a question on an AS core 1 paper. Im not really sure what its asking me to do.

Find constants a and b, such that for all values of x:

x^2 + 4x + 14 = (x + a)^2 +b - Jan 1st 2008, 11:32 PMwingless
$\displaystyle x^2 + 4x + 14 = (x+a)^2 + b$

$\displaystyle x^2 + 4x + 14 = \underbrace{x^2 + 4x + 4}_{\text{a square}} + 10 = (x+2)^2 + 10$

$\displaystyle (x+2)^2 + 10 = (x+a)^2 + b$

And as you see here, we found $\displaystyle a = 2$ and $\displaystyle b = 10$ - Jan 2nd 2008, 12:09 AMmr fantastic
- Jan 2nd 2008, 02:51 AMIsomorphism

There is a need to understand the statement correctly. I think Mr.F's suggestion should be followed because it is precise and complete.

For two polynomials p and q of finite degree, if $\displaystyle \forall x\in \mathbb{R}$ p(x) = q(x) holds, then they are identical polynomials.*Idea:*

This idea is important because they are embedded within the definition of a polynomial. For polynomials to be well defined, we need a condition to know when they are same.

So using this idea shows that you have understood the definition of polynomials.

So write the statement:

$\displaystyle \forall x \in \mathbb{R} ,x^2 + 4x + 14 = (x + a)^2 +b = x^2 + (2a)x + (a^2 + b)$

This means the polynomials are identical and hence $\displaystyle 2a = 4$ and $\displaystyle a^2+b = 14$

This method says this must be the**ONLY**solution.

If you follow the eyeballing method, there is an additional question you have to ask yourself. and that is

: "Oh alright, I can see one solution clearly, but what if there are other solutions? What if, for some other values of a and b, you get the same equation?".*Introspection*

If you can see why there cant be other solutions, then this method is fine too. But can you? ;) - Jan 2nd 2008, 02:58 AMSean12345
Isomorphism is right, however for the core 1 AS paper they like you to solve by inspection (weird I know but true). Indeed if the question proceeds to be more taxing then i would personally use the latter method.

- Jan 2nd 2008, 03:11 AMwingless
There can't be another solution because in order to get 4x, a must be 2. Then b has only one value too.

I just thought that it was a 8th-9th grade question and he wouldn't know these things about polynomials. Surely mr fantastic's solution is more general and can be applied to other questions like this. ;) - Jan 2nd 2008, 05:01 AMIsomorphism
Yes ,of course, wingless.I have nothing against your solution.

__My opinions:__

I just thought I will educate the poster about the idea. He has already thanked you, so most likely he will not even read this. But in case some one else is reading this thread, I thought it will help them out. The different ideas for a solution and learning the pitfalls of junior class methods are important processes in learning.I wanted him to think, thats all.

But in the hindsight, I think you are right, probably it is bad to over-instruct too :p

Quote:

Originally Posted by**Sean12345**

- Jan 2nd 2008, 05:07 AMbobak

It is the first exam british Students doing Advance Level math take it is required by all British universities for undergraduate degrees, but many require additional test.

here are some sample papers

AQA GCE A/AS Mathematics

Edexcel : Qualifications : GCE including applied subjects - Jan 2nd 2008, 05:11 AMIsomorphism
- Jan 2nd 2008, 07:52 AMbobak
The question posted really isn't a proper Advance Level it more of a confidence booster question so most people should be able to solve it by inspection.

However the method i was taught for these problems was called "completing the square"

which is pretty much this.

$\displaystyle x^2 + ax + b = \left ( x + \frac{a}{2} \right )^2 - \left ( \frac{a}{2} \right )^2 +b $

**nugiboy**should be familiar with this method form doing GCSE maths.