Find the value of k so that the equation has (a) one solution, (b) two real solutions, (c) two imaginary solutions. x^2-2kx+k=0

Answer 1

Write down the quadratic formula for the solutions, which I assume you know.

(If you do not know what formula I mean, then let me know.)

x = (2k ± √(4k² - 4k))/2

The quantity (4k² - 4k) under the square root is called the "discriminant"

of the equation, because it discriminates among the three possibilities a), b), c).

If the discriminant is 0 then the whole square root is 0, and you get just one solution 2k/2.

If the discriminant is positive then the square root is a positive real number,

and you get two solutions, as indicated by the symbol ±.

If the discriminant is negative then the square root is a complex number

(although the option c uses the inaccurate term "imaginary").

For what follows it will be convenient to factor the discriminant into

4k(k-1)

So option a) is for those k satisfying

4k(k-1) = 0

which is k = 0 and k = 1.

Option b) is for those k satisfying

4k(k-1) > 0

That happens if either

k > 0 and k-1 > 0 (which simplifies into k > 1), or

k < 0 and k-1 < 0 (which simplifies into k < 0).

For option c) you can do the same kind of analysis as for b), or simply list the

remaining numbers: 0 < k < 1.