Polynomial Perfect Square

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Johnc227

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From Advanced Maths Core 1 & 2. (Neil and Quadling), Misc Exercise 9, P 136, No. 3.

Let p(x) = x^2 -6x -3 and q(x) = x^2 - 2x + 4
(a) Calculate p(x) - q(x) and p(x)q(x). I had no problems here.

The polynomial p(x)+aq(x), where a is a constant, is a perfect square.
(b)* Calculate two possible values of a.
(the asterisk * is included in the question I do not know what it refers to. I have religiously worked through the book so far and must have missed something).

My thinking was that I needed to make the linear term cancel out and simultaneously make the leading quadratic coefficient and the constant become square numbers (terms).
If I make a =-3 then I can make the linear term become +6x but this does not help with the other terms. Also why would there be more than one way to make +6x from -2x?)

The answers in the book are; 3 and -4/3.

I do not understand the answers.
If I times q(x) by +3, the leading coefficient would become 4x^2 and the constant becomes 9 which are both square terms but the linear term becomes 12x.
If I times q(x) by -4/3 it all becomes messy to me.
(and I thought this would be an easy chapter).

I am grateful to Subhotosh Khan who helped me previously and to anyone looking at this.

John C
 
"The polynomial p(x)+aq(x), where a is a constant, is a perfect square." brings to mind this formula: (b + c)2 = b2 + 2bc + c2 (not using a and b to avoid confusion with your a). So you should find a, such that the resulting polynomial looks like the right side of the formula.
 
[MATH](ux + v)^2 = u^2x + 2uvx + v^2.[/MATH]
[MATH]x^2 - 6x - 3 + a(x^2 - 2x + 4) = (a + 1)x^2 + (-\ 2a - 6) + 4a - 3.[/MATH]
Equate coeffecients.

[MATH]u^2 = a + 1,\ 2uv = - (2a + 6),\ \text { and } v^2 = 4a - 3.[/MATH]
You have 3 unknowns and three equations.

[MATH]\therefore v = \pm \sqrt{4a - 3}.[/MATH]
[MATH]\text {CASE I: } v = \sqrt{4a -3}.[/MATH]
[MATH]\therefore 2uv = -\ (2a + 6) \implies uv = - \ (a + 3) \implies \text{WHAT?}[/MATH]
EDIT: Basically, my answer follows the same line of thought as lev's, but I have taken it a bit further. Equating coeffecients is a useful tool to have in your toolbox.
 
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Johnc227 said:
(b)* Calculate two possible values of a.
(the asterisk * is included in the question I do not know what it refers to. I have religiously worked through the book so far and must have missed something).
Click to expand...
An asterisk before an exercise often means "this is a challenging problem, beyond what you have been explicitly taught", or something like that. Look in the introduction to your book (maybe a section called "To the student" or "How to use this book") to find explanations of such notation.
 
There are typos in my prior post. Sloppy checking. I apologize.

[MATH](ux + v)^2 = u^2x^2 + 2uv + v^2.[/MATH]
[MATH](a + 1)x^2 + (-\ 2a - 6)x + (4a - 3).[/MATH]
Now you can equate coefficients.

Also, you should consider whether it is possible that 4a - 3 = 0.
 
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Lev, Indeed.

Thank you very much.

Fixed now.
 
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Thank you Lev, Jeff and Dr Peterson.
My maths is not strong enough at the moment to quickly follow and I need time to digest what you have said but I am very grateful.
I am off for a few days but will report back.
Thanks again.
 
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