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. |
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.... |
Originally Posted by camk
3X^2+12X+5 = P(X+Q)^2+R
(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 |
LOL - need to be quicker next time.
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Originally Posted by dsmith
LOL - need to be quicker next time.
And I let the lad work out the last bit for himself... you have just cost him a grade for cheating! ;) |
And I let the lad work out the last bit for himself... you have just cost him a grade for cheating! ;) Alcazar |
I take it your son is at university ?
Really, that all means nothing to me, havent go a clue.. |
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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