Question

If $x_1$ and $x_2(x_1>x_2)$ are the roots of the equation $\sqrt{(6-x)}(3^{x^2-7.2x+3.9}-9\sqrt{3})=0$ then calculate $x_1-\frac{1}{x_2}$

Answer 1

We have $\sqrt{(6-x)}(3^{x^2-7.2x+3.9}-9\sqrt{3})=0$
This equation is defined for
$6-x\ge 0i.ex\le 6$ ....(1)

The given equation is equivalent to the following equations
$\sqrt{6-x}=0$ and $3^{x^2-7.2x+3.9}-9\sqrt{3}=0$
$\therefore x_1=6$ and $3^{x^2-7.2x+3.9}=3^{2.5}$
$\Rightarrow x^2-7.2x+3.9=2.5$

$\Rightarrow x^2-7.2x+1.4=0$
We find that $x_2=\frac{1}{5}$ and $x_3=7$
Hence solution of the original equation are
$x_1=6$ and $x_2=\frac{1}{5}$ (Since, for $x=7$, the given equation is not defined )

Now, $x_1-\frac{1}{x_2}=6-\frac{1}{1/5}=1$