Question

How many integer values are there for k for which 4x 2 + kxy−9y 2 is factorable?

Answer 1

are you sure about the sign of y=9y^2?

Answer 2

yes. the answer key says 9.

Answer 3

Suppose you can factor it so that you get something like
\[(2x+3y)(2x-3y)\]
Expanding, you get
\[4x^2+6xy-6xy-9y^2\\
4x^2-9y^2\]
So here, \(k=0\). That's one.

Since you're looking for something that's factorable, it looks like you'll have to list the all the factors of 4 and -9, then write them in this form
\[\bigg([\text{factor of 4}]x+[\text{factor of -9}]y\bigg)\bigg([\text{factor of 4}]x+[\text{factor of -9}]y\bigg)\]
Expand that, and \(k\) is the coefficient of \(xy\). Does that kinda make sense? This is one of those questions I'd be able to answer, but not really explain the reasoning well.

Answer 4

it makes sense to me. the Asker give out the answer k =9. I have no way to get there.

Answer 5

That's why I asked the sign of 9y^2, if it is +, quite easy to get the answer, but he said that it is - . :(

Answer 6

I think it actually helps in this case. It tells you that the factored form has
\[(\cdot +\cdot)(\cdot-\cdot)\]
(The dots are factors)

Answer 7

note our highest-degree terms are both perfect squares -- that's our hint

Answer 8

nope. it's is 4x^2 -9y^2

Answer 9

OOPS

Answer 10

still dont get it ._.

Answer 11

Factors of 4: 1, 2, 4
Factors of -9: (±) 1, 3, 9 ... ... (If one is positive, the other has to be negative)

How many possible sums can you make here between two numbers, given the condition that one of the factors of 9 has to be negative?

Answer 12

Actually, that condition may not matter... but I'll leave it there just in case.

Answer 13

The factors of -36 are:
1. -36(1)
2. 36(-1)
3. -18(2)
4. 18(-2)
5. -12(3)
6. 12(-3)
7. -9(4)
8. 9(-4)
9. -6(6)

Answer 14

Add all of those pairs of factors and you will have the 9 possible values of k

Answer 15

So: k=-35, 35, -16, 16, -9, 9, -5, 5, 0

Answer 16

@Mertsj the final answer doesn't make sense to me at all. I think you just focus on factor the coefficient only, so ,k must be product of 4 and 9 . But it seems I misunderstand. Please, explain me

Answer 17

@Mertsj +10000 unfortunately I cannot give another medal

Answer 18

If k = -35 we have:
4x^2-36xy+1xy-9y^2=
4x(x-9y)+y(x-9y)=(x-9y)(4x+y) and so on for all the other examples.

Answer 19

@Mertsj got you, thanks

Answer 20

@Loser66 yw

Answer 21

it's fun when the person who makes question is me!! where is the Aaaasker!! @mippo12

Answer 22

then?

Answer 23

You brainy people need to go work on this one for awhile:
http://openstudy.com/study#/updates/51c0f775e4b0bb0a4360050a

Answer 24

@tcarroll010 \(k^2+144=c^2\implies (c+k)(c-k)=144\) so then we just look at integer factors of \(144\); we count only \(\lceil 15/2\rceil=8\) possible pairs of \((c+k)(c-k)\) and thus \(8\) possible \(k\) :-(

Answer 25

I have heard of beating a dead horse before...

Answer 26

@Mertsj I'm asking @tcarroll010 about using the discriminant to solve the question, the method, not just the silly final answer (which takes 0 skill)... you must really not know what a dead horse is

Answer 27

Since \(c-k,c+k\) are interchangeable presume \(c-k\le c+k\) and our argument works by symmetry:$$(c-k,c+k)\in\{(1,144),(2,72),(3,48),(4,36),(6,24),(8,18),(9,16),(12,12)\}$$Given that \(k=1/2[(c+k)-(c-k)]\) it follows that \((c+k)-(c-k)\) is even and therefore \(c+k,c-k\) are either both even or both odd; this allows us to 'weed out' pairs of factors which do not yield integral \(k\) and we are left with:$$(c-k,c+k)\in\{(2,72),(4,36),(6,24),(8,18),(12,12)\}$$now observe these yield \(k\in\{35,16,9,5,0\}\). For \(c+k\le c-k\) we then observe that \(k\in\{-35,-16,-9,-5,0\}\) and so in total we have \(9\) possible values of \(k\)