Albert1
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find:$x$
$x^2+\dfrac {9x^2}{(x-3)^2}=16$
$x^2+\dfrac {9x^2}{(x-3)^2}=16$
Albert said:thanks ! your solution is correct ( a bit tedious solution as you said )
I think you are very good at trigonometry:)
very good :)Pranav said:Solution without trigonometry:
The given equation can be written as
$$\left(x+\frac{3x}{x-3}\right)^2-\frac{2\cdot x\cdot 3x}{x-3}=16$$
$$\Rightarrow \left(\frac{x^2}{x-3}\right)^2-\frac{6x^2}{x-3}-16=0$$
Let $\frac{x^2}{x-3}=t$. Hence, we have:
$$t^2-6t-16=0$$
Solving we get, $t=8,-2$.
Case i), when $t=8$,
$$\frac{x^2}{x-3}=8 \Rightarrow x^2-8x+24=0$$
Clearly, the above equation has no solution as the discriminant is less than zero.
Case ii), when $t=-2$,
$$\frac{x^2}{x-3}=-2 \Rightarrow x^2+2x-6=0$$
Solving for x, we get, $x=-1+\sqrt{7},-1-\sqrt{7}$.