If (px+q)^2 = 9x^2 + kx + 16, what is the value of k?

max

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Jun 1, 2007
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If (px+q)^2 = 9x^2 + kx + 16, what is the value of k?

(a)3, (b)4, (c)12, (d)24, (e)30

trying to solve the problem, im getting kx = 2pq = (2)(3)(4) = 24. I compared with the answer in my book and it says k = 2pq = (2)(3)(4) = 24. What happend to x?

Can anyone explain?

thanks
 

stapel

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How did you get that 2pq equalled something with "x" in it? There is no "x" in either of p or q, that I can tell...?

It would be different, of course, if you were comparing kx and 2pqx, the middle term (rather than just the middle coefficient) of the expanded binomial, but I'm afraid I'm not able to figure out the steps between the exercise and your ending point.

Please reply showing the rest of your work. Thank you! :D

Eliz.
 

max

Junior Member
Joined
Jun 1, 2007
Messages
50
If (px+q)^2 = 9x^2 + kx + 16, what is the value of k?

(a)3, (b)4, (c)12, (d)24, (e)30
I got it!

thanks

I missed the x in 2pq

So (px+q)^2 = p^2x^2 + 2pq(x!) + q^2 = 9x^2 + kx + 16

p^2 has to = 9 and q^2 has to = 16. P can be +/- 3 and q +/- 4.

Since the middle terms of p^2x^2 + 2pqx + q^2 and 9x^2 + kx + 16 must be equal, I should get kx = 2pqx = 2(3)(4)x = 24x so k equals 24.

I guess also I could use the negative numbers to get the same answer. kx = 2(-3)(-4)x = 24x. The answer is correct as long as both (3) and (4) are both positive or both negative.
 
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