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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

find:$x$

$x^2+\dfrac {9x^2}{(x-3)^2}=16$

$x^2+\dfrac {9x^2}{(x-3)^2}=16$

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

find:$x$

$x^2+\dfrac {9x^2}{(x-3)^2}=16$

$x^2+\dfrac {9x^2}{(x-3)^2}=16$

- Admin
- #2

- Feb 14, 2012

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My workable but a bit tedious solution:

$1+\left( \dfrac{3}{x-3} \right)^2=\left( \dfrac{4}{x} \right)^2$

Let $\dfrac{3}{x-3}=\tan y$, $\therefore\dfrac{4}{x}=\dfrac{4\tan y}{3(1+\tan y)}$ and the equation above becomes

$1+\tan^2 y=\left( \dfrac{4\tan y}{3(1+\tan y)} \right)^2$, which after simplification, we get

$\sec y= \dfrac{4\tan y}{3(1+\tan y)} $.

By rewriting $\sec y= \dfrac{4\tan y}{3(1+\tan y)} $ as another equation that involves only $\sin y$ and $\cos y$, we obtain

$3(\sin y+\cos y)=4\sin y \cos y$

Solving it by squaring both sides of the equation we have

$4\sin^2 2y-9\sin 2y-9=0$

$\sin 2y=3$ or $\sin 2y=-\dfrac{3}{4}$ (Since $\sin 2y\le 1$, $\sin 2y=-\dfrac{3}{4}$ is the only solution.)

$\sin 2y=-\dfrac{3}{4}$ tells us $2y$ lies in the third and fourth quadrant.

For $2y$ that lies in the third quadrant: | For $2y$ that lies in the fourth quadrant: |

$\tan 2y=\dfrac{3}{\sqrt{7}}$ | $\tan 2y=-\dfrac{3}{\sqrt{7}}$ |

$\dfrac{2\tan y}{1-\tan^2 y}=\dfrac{3}{\sqrt{7}}$ Solving the equation for $\tan y$ and takes only the negative result, we see that $\tan y=\dfrac{4-\sqrt{7}}{3}$ By replacing $\tan y=\dfrac{4-\sqrt{7}}{3}$ into $\dfrac{3}{x-3}=\tan y$, we get the value of $x$ as $x=\sqrt{7}-1$ | $\dfrac{2\tan y}{1-\tan^2 y}=-\dfrac{3}{\sqrt{7}}$ Solving the equation for $\tan y$ and takes only the negative result, we see that $\tan y=\dfrac{4-\sqrt{7}}{3}$ By replacing $\tan y=\dfrac{\sqrt{7}-4}{3}$ into $\dfrac{3}{x-3}=\tan y$, we get the value of $x$ as $x=-\sqrt{7}-1$ |

- Thread starter
- #3

- Jan 25, 2013

- 1,225

thanks ! your solution is correct ( a bit tedious solution as you said )

I think you are very good at trigonometry

- Admin
- #4

- Feb 14, 2012

- 3,931

Thanks for the compliment,thanks ! your solution is correct ( a bit tedious solution as you said )

I think you are very good at trigonometry

- Nov 4, 2013

- 428

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, $x=-1+\sqrt{7},-1-\sqrt{7}$.

- Thread starter
- #6

- Jan 25, 2013

- 1,225

very goodSolution 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, $x=-1+\sqrt{7},-1-\sqrt{7}$.