Question

$$\sqrt { x } + \sqrt { y } = \sqrt { a }$$

Solve for a

$a=\left(\sqrt{x}+\sqrt{y}\right)^{2}$
$x\geq 0\text{ and }y\geq 0$

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Solve for x

$x=\left(\sqrt{y}-\sqrt{a}\right)^{2}$
$y\geq 0\text{ and }a\geq 0\text{ and }-\left(\sqrt{y}-\sqrt{a}\right)\geq 0$

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Solve for a (complex solution)

$a=\left(\sqrt{x}+\sqrt{y}\right)^{2}$
$\left(x=0\text{ and }y=0\right)\text{ or }arg(\sqrt{x}+\sqrt{y})<\pi $

Solve for x (complex solution)

$x=\left(\sqrt{y}-\sqrt{a}\right)^{2}$
$y=a\text{ or }arg(\sqrt{y}-\sqrt{a})\geq \pi $