Mathematics - Quadratic Equation Question with Solution | TestHub

If $a=2$, the equation becomes $2x-2=0$, which gives $x=1$. So, A is correct. If $a=4$, the equation is $2x^2-2=0$, so $x^2=1$, giving roots $x=1$ and $x=-1$. So, B is incorrect. If $a\in (2,4)$, the coefficient of $x^2$ is $a-2>0$ and the constant term is $-2<0$. Since the product of the leading coefficient and the constant term is negative, $AC<0$, the roots are real and of opposite signs. So, C is correct. If $a\in (4,\infty )$, the discriminant $D=(a-4{)}^2-4(a-2)(-2)=(a-4{)}^2+8(a-2)=a^2-8a+16+8a-16=a^2$. The roots are $x=\frac{(a-4)\pm a}{2(a-2)}$. The roots are $x_1=\frac{a-4+a}{2(a-2)}=\frac{2a-4}{2(a-2)}=1$ and $x_2=\frac{a-4-a}{2(a-2)}=\frac{-4}{2(a-2)}=\frac{-2}{a-2}$. For $a\in (4,\infty )$, $a-2>0$, so $x_1=1$ (positive) and $x_2=\frac{-2}{a-2}$ (negative). Thus, one root is positive and the other is negative. So, D is incorrect.