Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=x^2-(3k-1)x+2k^2-2k$. For both roots to be greater than 1, three conditions must be met: (1) Discriminant $D\ge 0$, (2) $f(1)>0$, and (3) Vertex $x$ -coordinate $-b/(2a)>1$. (1) $D=(3k-1{)}^2-4(2k^2-2k)=9k^2-6k+1-8k^2+8k=k^2+2k+1=(k+1{)}^2\ge 0$, which is true for all real $k$. (2) $f(1)=1-(3k-1)+2k^2-2k=2k^2-5k+2>0$, which factors as $(2k-1)(k-2)>0$, so $k\in (-\infty ,1/2)\cup (2,\infty )$. (3) $-b/(2a)=(3k-1)/2>1\Rightarrow 3k-1>2\Rightarrow 3k>3\Rightarrow k>1$. Combining these conditions, we find the intersection of $k\in \mathbb{R}$, $k\in (-\infty ,1/2)\cup (2,\infty )$, and $k\in (1,\infty )$. The intersection is $k\in (2,\infty )$.