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-   -   Maths homework help !! (https://www.scoobynet.com/non-scooby-related-4/371294-maths-homework-help.html)

camk 16 October 2004 05:36 PM

Maths homework help !!
 
Guys,
The lads homework is getting to the edge of my knowledge, can anyone help.

Need to get the values of P,Q & R.

3X^2+12X+5 = P(X+Q)^2+R

where ^2 is the preceeding number squared. Obvioulsy we need to see how to work it out and not just the answers.

Thanks very much in advance.

ajm 16 October 2004 06:09 PM

Its been a while but from memory you need to make the right hand side of the equation match the left hand side e.g. in the form

(...)x^2 + (...)x + (...)

If you expand out the right hand side and assemble in that form you get:-

(P)x^2 + (2QP)x + (PQ^2 + R)

Now compare with the left hand side....

Let the numbers of X^2 be equal: P = 3

Let the numbers of x be equal each side: 2QP = 12

Let the remainder be the same: PQ^2+R = 5

Substitute in and you should get your answer....

dsmith 16 October 2004 06:11 PM


Originally Posted by camk
3X^2+12X+5 = P(X+Q)^2+R

we know (or should ;) ) that
(a+b)^2 = a^2 + 2ab + b^2

so
(X+Q)^2 = X^2 + 2QX + Q^2

so if
3X^2+12X+5 = P(X+Q)^2+R
= P(X^2 + 2QX + Q^2) + R
= PX^2 + 2QPX + PQ^2 + R

Then
P=3 (from X^2 terms)
12=2QP so Q= 2 (from X terms)
PQ^2 + R = 5 so R= -7 (from the non-x bits)

Deano

dsmith 16 October 2004 06:12 PM

LOL - need to be quicker next time.

ajm 16 October 2004 06:18 PM


Originally Posted by dsmith
LOL - need to be quicker next time.

:D

And I let the lad work out the last bit for himself... you have just cost him a grade for cheating! ;)

alcazar 16 October 2004 10:22 PM


And I let the lad work out the last bit for himself... you have just cost him a grade for cheating! ;)
Tell 'em his name's Harry, Edexcel will hush it up sooooo fast:D:

Alcazar

yoza 16 October 2004 11:15 PM

I take it your son is at university ?

Really, that all means nothing to me, havent go a clue..

camk 18 October 2004 08:51 PM

no he's doing A - Level maths at school. Thanks for the help lads I knew Scoobynet would assist. I just said I'm not sure but one of the lads will :D


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