Solving equation with fraction

Question

I don't understand how to get from the second to the third step in this equation:

$ - \frac { \sqrt { 2 - x ^ { 2 } } - x \left( \frac { - x } { \sqrt { 2 - x ^ { 2 } } } \right) } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \sqrt { 2 - x ^ { 2 } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \frac { 2 - x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { 2 } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 3 } } $

Why can we just add $ 2 - x ^ { 2 } $ in the numerator? Step 1 to step 2, as well as step 3 to step 4 is clear to me.

Answer 1

You multiply by $1$ in a fancy fashion:

$$\sqrt{2-x^2}=\sqrt{2-x^2}\cdot 1=\sqrt{2-x^2}\cdot \frac{\sqrt{2-x^2}}{\sqrt{2-x^2}}=\frac{(\sqrt{2-x^2})^2}{\sqrt{2-x^2}}=\frac{2-x^2}{\sqrt{2-x^2}}$$

This of course only holds if $2-x^2\neq 0$, as otherwise also $\sqrt{2-x^2}=0$ and the last expression(and the intermediate ones) would be ill-defined.

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Note, that his holds for any $a\in (0,\infty)$ by the same method as above, i.e.

$$\sqrt{a}=\sqrt{a}\cdot 1=\sqrt{a}\cdot\frac{\sqrt{a}}{\sqrt{a}}=\frac{(\sqrt{a})^2}{\sqrt{a}}=\frac{a}{\sqrt{a}}$$

Answer 2

Hint : Note that $2-x^2 = (\sqrt{2-x^2}) * (\sqrt{2-x^2})$ and $(\sqrt{2-x^2}) = \frac{2-x^2}{(\sqrt{2-x^2})}$

Answer 3

Is because for any $y>0$ $$ \sqrt{y}=\frac{y}{\sqrt{y}}. $$

Answer 4

It's simply that for $a>0$, $$\sqrt{a} = \sqrt{a}\cdot\underbrace{\left(\frac{\sqrt{a}}{\sqrt{a}}\right)}_1$$ $$=\frac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{a}}$$ $$=\frac{(\sqrt{a})^2}{\sqrt{a}}$$ $$=\frac{a}{\sqrt{a}}$$ In your case, "$a$" is $2-x^2$.

Answer 5

Because for $a > 0$ we know $\sqrt a = \frac a{\sqrt a}$ so $\sqrt{2 -x^2} = \frac {2-x^2}{\sqrt{2-x^2}}$ (assuming that $2 - x^2 > 0$ which must be the case as you have had $\sqrt{2 - x^2}$ in the denominator elsewhere in the equation).