Question

Help please!!....For what value of "k" does the equation kx^2 - 2kx + k - 1=0 not have solutions in the real number set?

Answer 1

\[{p(x)=kx^2-2kx+k-1}\] let !

Answer 2

for the value where k in in closed form

Answer 3

factors of -1 = \(\pm 1\)

Answer 4

p(-1)=\(\large{k(-1)^2-2k(-1)+k-1=0}\)
\(\large{4k-1=0}\)
\(\large{k=\frac{1}{4}}\)

Answer 5

i am just confused ...

Answer 6

As long as \(k\geqslant0\), the polynomial will have real solutions.

Answer 7

Hint:

\[\Large ax^2+bx+c=0\]

will have nonreal solutions when

\[\Large b^2 - 4ac < 0\]

Answer 8

make sure \(b^2-4ac\geq 0\) in this case \(a=k, b=-2k, c=(k-1)\)

Answer 9

thanks for all the help guys!