Question

x^2 + kx - 8 can be factored to (x + a)(x + b) with a and b as integers. What are 3 possible values of k?

Answer 1

If a quadratic can be factored with integer factors, the discriminant D must be a perfect square. In the equation ax^2 + bx + c, the discriminant D is defined as: b^2 - 4ac. Therefore, for the equation x^2 + kx - 8, the discriminant D = k^2 - (4)(1)(-8) = k^2 + 32

Answer 2

k^2 + 32 must be a perfect square.
Can you come up with a few possible values of k?

Answer 3

I know one would be -7, but I'm not sure how I got it.

Answer 4

The lowest value of k^2 is zero. It cannot be negative. It is always positive or zero.
So when you add k^2 and 32, the result will always be greater than or equal to 32. And we want the result to be a perfect square. What is perfect square greater than 32?

Answer 5

To find the first few perfect squares, square, 1, 2, 3, 4, 5, .....
First few perfect squares:
1, 4, 9, 16, 25, 36, 49, 64, 91, 100, 121, 144, 169, 196, 225, 256, ....

Answer 6

36.

Answer 7

Yes. k^2 + 32 = 36
k^2 = 4
k = -2, 2
So you got two values of k that will make k^2 + 32 a perfect square.

Answer 8

Okay, thank you.

Answer 9

Let us keep trying other perfect squares.
The on after 36 is 49
k^2 + 32 = 49
k^2 = 17.
But this gives a non-integer value for k. So we will skip this one.
Try others.

Answer 10

I have a typo in my earlier reply where I list the first few perfect squares. It should be 81 and not 91.

Answer 11

If we try 81, we have k^2 + 32 = 81
k^2 = 81 - 32 = 49
k = -7. 7
we have two more.

So k = -2, 2, -7, 7.

Answer 12

Thank you! I think I get it. I had originally gotten -7 as one answer.

Answer 13

If you put back any one of the k values we found you will find you will be able to factor them nicely. For example, with +2 we get: x^2 + 2x - 8 = (x + 4)(x - 2)

Answer 14

You are welcome.