Maths homework help !!
#1
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.
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.
#2
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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....
(...)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....
#3
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
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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!
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