Answer:
Discriminant
[tex]b^2-4ac\quad\textsf{when}\:ax^2+bx+c=0[/tex]
[tex]\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}[/tex]
[tex]\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}[/tex]
[tex]\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}[/tex]
Given equation: [tex]kx^2+2x+(1-k)=0[/tex]
[tex]\implies a=k,\:b=2,\:c=(1-k)[/tex]
[tex]\begin{aligned}\implies b^2-4ac & =(2)^2-4(k)(1-k)\\ & =4-4k(1-k)\\ & =4-4k+4k^2\\ & = 4k^2-4k+4 \end{aligned}[/tex]
Complete the square:
[tex]\begin{aligned}\implies 4k^2-4k+4 & =4(k^2-k+1)\\ & = 4\left[k^2-k+\left(\dfrac{-1}{2}\right)^2+1-\left(\dfrac{-1}{2}\right)^2\right]\\ & = 4\left[\left(k-\dfrac12\right)^2+\dfrac34\right]\\ & = 4\left(k-\dfrac12\right)^2+3\end{aligned}[/tex]
[tex]\textsf{As }\left(k-\dfrac12 \right)^2 \geq 0 \implies 4\left(k-\dfrac12 \right)^2+3\geq 3[/tex]
Therefore, the roots are always real.
