Need help with quadratics (revised)?

Question

If one root of the quadratic equation x^2-kx-2=0 is 2, find the value of k and the other root. here is what i have so far. am i on the right track.

sorry about the other question the known root is 2

Answer 1

x^2 - kx - 2 = 0
One root is 2.
To get the value of k, put x = 2

2^2 - 2k - 2 = 0
4 - 2 - 2k = 0
2 - 2k = 0
-2k = -2
2k = 2
k = 1

Now the equation is x^2 - x - 2 = 0
x^2 - x - 2 = 0
x^2 - 2x + x - 2 = 0
x(x - 2) + 1(x - 2) = 0
(x - 2)(x + 1) = 0
x = 2, -1

We already know that 2 is a root
So -1 is the other root

Answer 2

if one root is 2 then;replace all x by 2 you will have:
(2)^2 - 2k - 2 = 0
4 - 2k - 2 = 0
2 - 2k = 0
2 = 2k
k = 1 replace k in the original problem.

x^2 - (1)x - 2 = 0 , or x^2 - x - 2 = 0
then factor: ( x - 2 ) ( x + 1 ) = 0
set : ( x - 2 ) = 0 , then x = 2 ( one of the root, your clue)
set : ( x + 1 ) = 0 , then x = -1 ( the other root)

hope this can help.

Answer 3

One root is 2 (given).
Let us suppose that the other is a.
The sum of the roots is k, and the product of the roots is -2.
Therefore a+2 = k
and 2a = -2
Hence a = -1 and k = 1.

Answer 4

One root is 2. So, use that value as x.
2^2-k(2)-2=0
k=-1

To find the other root, put in the value, k=-1. x^2+x-2=0
Therefore, (x+2)(x-1)=0
x=-2 and x=1

The other root would be 1.

I hope this helps. :)

Answer 5

Substitue x = 2 in the eqn to kind the value of k = 1
Now you can find the other root = -1