Mathematics - Quadratic Equation Question with Solution | TestHub

For the quadratic equation $(a-1)x^2-(a+1)x+(a-1)=0$, for it to be a quadratic, $a-1\ne 0⟹a\ne 1$. For real roots, the discriminant $D\ge 0$. $D=(-(a+1){)}^2-4(a-1)(a-1)=(a+1{)}^2-4(a-1{)}^2=(a^2+2a+1)-4(a^2-2a+1)=-3a^2+10a-3$. So, $-3a^2+10a-3\ge 0⟹3a^2-10a+3\le 0$. The roots of $3a^2-10a+3=0$ are $a=\frac{10\pm \sqrt{100-36}}{6}=\frac{10\pm 8}{6}$, which are $a=1/3$ and $a=3$. Thus, $a\in [1/3,3]$. Combining with $a\ne 1$, we get $a\in [1/3,1)\cup (1,3]$. Let $f(x)=(a-1)x^2-(a+1)x+(a-1)$. The condition that one root is greater than 2 and the other root is less than 1 means that $1$ and $2$ lie between the roots. This implies that $(a-1)f(1)<0$ and $(a-1)f(2)<0$. First, $f(1)=(a-1)-(a+1)+(a-1)=a-3$. So $(a-1)(a-3)<0⟹1<a<3$. Second, $f(2)=(a-1)(4)-(a+1)(2)+(a-1)=4a-4-2a-2+a-1=3a-7$. So $(a-1)(3a-7)<0⟹1<a<7/3$. Both conditions must hold, so we need $a\in (1,3)$ and $a\in (1,7/3)$. The intersection is $a\in (1,7/3)$. This interval is consistent with $a\in [1/3,1)\cup (1,3]$ and $a\ne 1$. Therefore, the range of $a$ is $(1,7/3)$.