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 a 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.

$y=1$ satisfies the equation.

Since $3^x$ is positive and $3^x\ge 3^0$, we have $y\ge 1$. The product of the roots is $1\times y>1$.

$$\begin{array}{rl}& \Rightarrow \frac{1}{2a-4}>1\\ & \Rightarrow 2a-4<1\\ & \Rightarrow a<\frac{5}{2}\end{array}$$

The sum of the roots is $\frac{2a-3}{2a-4}>1$.

$$\begin{array}{rl}& \Rightarrow \frac{(2a-3)-(2a-4)}{2a-4}>0\\ & \Rightarrow \frac{1}{2a-4}>0\\ & \Rightarrow a>2\\ & \Rightarrow 2<a<\frac{5}{2}\end{array}$$