If $x = y$, $p$ = what?

Question

It is given that,

$$ \frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p $$

If $x=y$, $p$ =?

Answer 1

We have: $\dfrac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p$

Let's replace $x$ with $y$:

$\Rightarrow p=\dfrac{\sqrt{2y+3y}+\sqrt{2y-3y}}{\sqrt{2y+3y}-\sqrt{2y-3y}}$

$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{-y}}{\sqrt{5y}-\sqrt{-y}}$

$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}-\sqrt{y}\hspace{0.5 mm}i}$

Then, let's multiply this fraction by the complex conjugate of its denominator:

$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}-\sqrt{y}\hspace{0.5 mm}i}\cdot{\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}}$

$\hspace{9 mm} =\dfrac{\big(\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i\big)^{2}}{\big(\sqrt{5y}\big)^{2}-\big(\sqrt{y}\hspace{0.5 mm}i\big)6{2}}$

$\hspace{9 mm} =\dfrac{5y+2\cdot{\sqrt{5y}}\cdot{\sqrt{y}\hspace{0.5 mm}i}+y\cdot{i^{2}}}{5y-y\cdot{i^{2}}}$

$\hspace{9 mm} =\dfrac{5y+2{\sqrt{5y^{2}}\hspace{0.5 mm}i}-y}{5y+y}$

$\hspace{9 mm} =\dfrac{4y+2{\sqrt{5y^{2}}\hspace{0.5 mm}i}}{6y}$

$\hspace{9 mm} =\dfrac{2y+2\sqrt{y^{2}}\sqrt{5}\hspace{0.5 mm}i}{3y}$

$\hspace{9 mm} =\dfrac{2y+2y\sqrt{5}\hspace{0.5 mm}i}{3y}$

$\hspace{9 mm} =\dfrac{2+2\sqrt{5}\hspace{0.5 mm}i}{3}$; $y\neq{0}$

Answer 2

$ \frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p $

Apply componendo and dividendo,

$ \frac{\sqrt{2x+3y}+\sqrt{2x-3y}+\sqrt{2x+3y}-\sqrt{2x-3y}}{\sqrt{2x+3y}+\sqrt{2x-3y}- \sqrt{2x+3y}+\sqrt{2x-3y}}=\frac{p+1}{p-1} $

$ \frac{\sqrt{2x+3y}}{\sqrt{2x-3y}} = \frac{p+1}{p-1} $

Put x=y

$ \frac{\sqrt{5y}}{\sqrt{-y}} = \frac{p+1}{p-1} $

Squaring both sides,

$5.(p-1)^2=-1(p+1)^2$

Sorry I have little mistake here, $5p^2-10p+5 = -p^2-2p-2$

$5p^2-10p+5 = -p^2-2p-1$

$6p^2-8p+6=0$

$3p^2-4p+3=0$

$D = b^2 - 4ac$

= $(-4)^2 - 4.3.3$ = 16 - 36 = - 20

$ p = \frac{-b \pm \sqrt D}{2a}$

= $\frac{-(-4) \pm \sqrt{-20}}{2.3}$

= $\frac{4 \pm \sqrt{20i^2}}{6}$

= $\frac{4 \pm 2i\sqrt5}{6}$

p = $\frac{2 + i\sqrt5}{3}, \frac{2 - i\sqrt5}{3}$

Edit-

In above formula D is discriminat. It is used when we can't factorise equation using factorisation method.

Here's the link when you can't find root using factorisation use discrimat method.

Answer 3

Since $x=y$, let's call them both $a$.

Then substituting, we have:

$$ p=\frac{\sqrt{5a}+\sqrt{-a}}{\sqrt{5a}-\sqrt{-a}} $$

If we multiply the top and bottom by the conjugate, $\sqrt{5a}-\sqrt{-a}$, we get: $$ p=\frac{5a-(-a)}{5a-2\cdot\sqrt{5a}\cdot\sqrt{-a}+(-a)} $$$$ p=\frac{6a}{4a-2\cdot\sqrt{-5a^2}}=\frac{6a}{4a-2\cdot a\cdot\sqrt{-5}} $$$$ p=\frac6{4-2\cdot\sqrt{5}\cdot\sqrt{-1}}=\frac3{2-\sqrt{5}\cdot i} [if a\neq 0] $$

This can be further simplified, but from here, a calculator can be used to solve the complex fraction.