What are the positive values for k for this equation? x^2 +x - k?

Question

How can you figure out the factoring when no numbers are involved?
Please help!

Answer 1

k can be a lot of different numbers ....
by assigning k a number, it is negative.... one positive factor and one negative factor..... +1 is the sum of those factors....
so if k = (first #) , then the factors are ...[ ]
(2, [-1,2])
(6, [-2,3])
(12[-3,4])
(20[-4,5])... the factors differ by one, the smaller being the negative... always gives you a +1 as your "B"


so you can use ....-1(n)(n+1)...just substitute numbers for "n"


[-6 * 7] = -42 and so on...

Answer 2

First off, you do not have an equation; you only have an expression. If the give expression is set =0, then you have an equation.

K must be a number such that its factors are consecutive figits.
Thus k = n(n+1) where n =>1
so possible values for k are 2 =1*2, 6 = 2*3, 12 = 4*3 etc
So x^+x-2 =(x+2)(x-1)
x^2 + x -6 = (x+3)(x-2)
x^2+x-12 = (x+4)(x=3)

So the possible values for k that make the expression factorable are infinite.

Answer 3

But I'll make a guess that you want to find pos values of k for which the expression has factors. We need numbers which have two factors differing by 1, such as
2*1=2
3*2=6
4*3=12
5*4=20
etc.

In other words, k=n(n+1) where n is a positive integer.

For example, if k=8*9 = 72,
we get
x^2 +x - 72 = (x+9)(x-8)

Answer 4

That's not an equation. It has to be equal to something to be an equation. Also, what is the underlying set: natural numbers, integers, real numbers, ...?

Your question cannot be answered as it is written.