camk

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

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

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

LOL - need to be quicker next time.

ajm

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

alcazar

iTrader: (2)

Tell 'em his name's Harry, Edexcel will hush it up sooooo fast:

Alcazar

yoza

I take it your son is at university ?

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

camk

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