1. ## Discriminant

Determine the value(s) of k for which x2 + (k-2)x - 2k = 0 has equal and real roots.

2. Hello, xxlvh!

Determine the value(s) of $k$ for which . $x^2 + (k-2)x - 2k \:= \:0$
has equal and real roots.

We're expected to know that a quadratic has equal roots if its discriminant is zero.
. . That is: . $b^2-4ac \:=\:0$

We have: . $a \:= \:1,\;b \:= \:k-2,\;c \:= \:\text{-}2k$

Therefore: . $(k-2)^2 - 4(1)(\text{-}2k) \:=\:0$ . . . . now solve for $k.$

3. seasons greetings. =)

Okay, so I had had simplified it down to k2-12k+4 = 0 theen
my final solution was k = 6, and -2.

4. Originally Posted by xxlvh
seasons greetings. =)

Okay, so I had had simplified it down to k2-12k+4 = 0 theen
my final solution was k = 6, and -2.
that is wrong. you expanded Soroban's expression incorrectly

(you can check your own solutions by plugging in the values you got for k and see if you get zero. you may want to plug them into Soroban's expression as a double check)