Question

x^2-2kx+1=0 has two distinct real roots. find the set of all possible values of K

Answer 1

Haha "oh. don't I love these 'k' questions"

Well first to start off with, we know we can utilize the discriminant. We have a quadratic with \(a,b,\) and \(c\) values of \(1,-2k,\) and \(1\) (respectively).
Let us utilize the discriminant \(D\):
\[\eqalign{
D&=b^2-4ac \\
&=4k^2-4(1)(1) \\
&=4k^2-4 \\
}\]
In order for there to be two distinct roots, we need \(D\gt 0\). So then we have:
\[\eqalign{
&4k^2-4\gt0 \\
&4k^2\gt4 \\
&k^2\gt1 \\
&k\gt\pm1 \\
}\]

So therefore, we obtain the set:
\[\{k\in R\phantom{.};\phantom{.}|k|\gt1\}\]
Or also we can have the set:
\[\{k\in R\phantom{.};\phantom{.}k\gt\pm1\}\]
Either one is acceptable I believe

Answer 2

thank :D

Answer 3

No prob!