Question

Set values of $K$ for which roots of the quadratic $x^2-(2K-1)x+K(K-1)=0$ are

• A. both less than $2$ is $K\in (2,\infty )$
• B. of opposite sign is $K\in (-\infty ,0)\cup (1,\infty )$
• C. of same sign is $K\in (-\infty ,0)\cup (1,\infty )$
• D. both greater than $2$ is $K\in (2,\infty )$

Answer 1

Answer: (C)

of same sign is $K\in (-\infty ,0)\cup (1,\infty )$

Expression : $x^2-(2K-1)x+K(K-1)=0$
$\Rightarrow$ $x^2-(K+K-1)x+K(K-1)=0$
$\Rightarrow$ $(x-K)(x-K+1)=0$
$\Rightarrow$ $x=Korx=K-1$
(A) If both roots are less than 2, then the bigger one of the two roots should be less than 2,
$\Rightarrow$ $K<2$ $\Rightarrow$ $K\in (-\infty ,2)$
(B) If both roots are of opposite sign, then their product should be negative,
$\Rightarrow$ $K(K-1)<0$
$\Rightarrow$ $K\in (0,1)$
(C) If both the roots are of the same sign, then their product should be positive,
$\Rightarrow$ $K(K-1)>0$
$\Rightarrow$ $K\in \{(-\infty ,0)\cup (1,\infty )\}$
(D) If both roots are greater than 2, then the smaller one of the two roots should be greater than 2,
$\Rightarrow$ $K-1>2$
$\Rightarrow$ $K>3$
$\Rightarrow$ $K\in (3,\infty )$
Hence, option (C) is the only correct choice.