Mathematics - Quadratic Equation Question with Solution | TestHub

Putting $3^x=y$, we have

$$(2a-4)y^2-(2a-3)y+1=0$$

This equation must have real solution

$\Rightarrow (2a-3{)}^2-4(2a-4)\ge 0$

$\Rightarrow 4a^2-20a+25\ge 0$

$\Rightarrow (2a-5{)}^2\ge 0$. This is true.

$y=1\text{ satisfies the equation }$

Since $3^x$ is positive and $3^x\ge 3^0,y\ge 1$

Product of the roots $=1\times y>1$

$\Rightarrow \frac{1}{2a-4}>1$

$\Rightarrow 2a-4<1\Rightarrow a<\frac{5}{2}$

Sum of the roots $=\frac{2a-3}{2a-4}>1$

$\begin{array}{ll}\Rightarrow & \frac{(2a-3)-(2a-4)}{2a-4}>0\\ & \end{array}$

$\Rightarrow \frac{1}{2a-4}>0\Rightarrow a>2\Rightarrow 2<a<\frac{5}{2}$