Question

Rationalise the denominator 1/5+root 2

Answer 1

Explanation:

Answer:

\frac{5-\sqrt{2}}{27}

27

5−

2

Explanation:

This kind of problem uses a sort of algebra trick.

To rationalize \frac{1}{5+\sqrt{2}}

5+

2

1

, multiply both numerator and denominator by what is called the conjugate radical -- the same expression but with opposite sign.

Note: this is really a clever name for 1, and multiplying by 1 does not change the value of an expression, it changes the way the expression looks.

Here goes:

\begin{gathered}\frac{1}{5+\sqrt{2}} \cdot \frac{5-\sqrt{2}}{5-\sqrt{2}}=\frac{5-\sqrt{2}}{25+5\sqrt{2}-5\sqrt{2}+2}\\=\frac{5-\sqrt{2}}{25+2}\\=\frac{5-\sqrt{2}}{27}\end{gathered}

5+

2

1

⋅

5−

2

5−

2

=

25+5

2

−5

2

+2

5−

2

=

25+2

5−

2

=

27

5−

2

See how the radical disappears from the denominator?

Answer 2

`(5-√(2))/(27)`

Explanation:

This kind of problem uses a sort of algebra trick.

To rationalize
`(1)/(5+√(2))`, multiply both numerator and denominator by what is called the conjugate radical -- the same expression but with opposite sign.

Note: this is really a clever name for 1, and multiplying by 1 does not change the value of an expression, it changes the way the expression looks.

Here goes:

`(1)/(5+√(2)) \cdot (5-√(2))/(5-√(2))=(5-√(2))/(25+5√(2)-5√(2)+2)\\=(5-√(2))/(25+2)\\=(5-√(2))/(27)`

See how the radical disappears from the denominator?