# To rationalize the denominator of the expression \frac{1}{\sqrt{2}+x}

To rationalize the denominator of the expression $$\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Bromodermgs
Calculation:
The denominator of a rational expression can be rationalized by removing the irrational term from it.
To remove the irration term, the numerator and denominator of the expression are multiplied by the conjugate of the denominator.
Conjugate of an expression $$\displaystyle\sqrt{{{a}}}+\sqrt{{{b}}}$$ is given as $$\displaystyle\sqrt{{{a}}}-\sqrt{{{b}}}$$ and vice-versa.
So the conjugate of the denominator is given as $$\displaystyle\sqrt{{{2}}}-{x}$$
Multiplying the numerator and the denominator with the conjugate,
$$\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}\times{\frac{{\sqrt{{{2}}}-{x}}}{{\sqrt{{{2}}}-{x}}}}$$
$$\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{\sqrt{{{2}}}-{x}}}{{{\left(\sqrt{{{2}}}+{x}\right)}{\left(\sqrt{{{2}}}-{x}\right)}}}}$$
$$\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{\sqrt{{{2}}}-{x}}}{{{\left(\sqrt{{{2}}}\right)}^{{{2}}}-{\left({x}\right)}^{{{2}}}}}}{i}.{e}.{\left({a}+{b}\right)}{\left({a}-{c}\right)}={a}^{{{2}}}-{b}^{{{2}}}$$
$$\displaystyle{\frac{{{1}}}{{\sqrt{{{2}}}+{x}}}}={\frac{{\sqrt{{{2}}}-{x}}}{{{2}-{x}^{{{2}}}}}}$$ (Since square and square root are opposite operations, they cancel out).