Question

16. Given that k is a real constant such that 0 < k < 1. Show that the roots of the equation kx2 + 2x + (1 – k) = 0, are С (i) Always real (ii) Always negative​

Answer 1

Discriminant

`b^2-4ac\quad\textsf{when}\:ax^2+bx+c=0`

`\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}`

`\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}`

`\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}`

Given equation:
`kx^2+2x+(1-k)=0`

`\implies a=k,\:b=2,\:c=(1-k)`

`\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}`

Complete the square:

`\begin{aligned}\implies 4k^2-4k+4 & =4(k^2-k+1)\\ & = 4\left[k^2-k+\left((-1)/(2)\right)^2+1-\left((-1)/(2)\right)^2\right]\\ & = 4\left[\left(k-\frac12\right)^2+\frac34\right]\\ & = 4\left(k-\frac12\right)^2+3\end{aligned}`

`\textsf{As }\left(k-\frac12 \right)^2 \geq 0 \implies 4\left(k-\frac12 \right)^2+3\geq 3`

Therefore, the roots are always real.