Question

rationalise the denominator 7-3 root 2 by 7+3 root 2

Answer 1

Hey there!

Give radical values to rationalise the terms "7 - 3 _/2 / 7 + 3 _/2" :

Multiply this by a conjugated term in order to rationalise it easily:

$\bf{\dfrac{(7 - 3 \sqrt{2}) (7 - 3 \sqrt{2})}{(7 + 3 \sqrt{2}) (7 - 3 \sqrt{2})}}$

Apply the exponential rule and perfect square formula that is (respectively):

$\boxed{\bf{a^b \times a^c = a^{b + c}}}$

AND

$\boxed{\bf{(a - b)^2 = a^2 - 2ab + b^2}}$

Here, a = 7  and  b = 3 _/2.

Therefore,

$\bf{\dfrac{7^2 - 2 \times 7 \times 3 \sqrt{2} + (3 \sqrt{2})^2}{(7 + 3 \sqrt{2}) (7 - 3 \sqrt{2})}}$

$\bf{\dfrac{49 - 42 \sqrt{2} + 18}{(7 + 3 \sqrt{2}) (7 - 3 \sqrt{2})}}$

$\bf{\dfrac{67 - 42 \sqrt{2}}{(7 + 3 \sqrt{2}) (7 - 3 \sqrt{2})}}$

Now, Rationalise the denominator to obtain the final value and the answer for this query:

Applying the basic principles for Two squares formula that is;  

$\bf{(a + b) (a - b) = a^2 - b^2}$

Therefore, Here,    a = 7   and   b = 3 _/2.

$\bf{\dfrac{67 - 42 \sqrt{2}}{7^2 - (3 \sqrt{2})^2}}$

$\bf{\dfrac{67 - 42 \sqrt{2}}{49 - 18}}$

$\bf{\dfrac{67 - 42 \sqrt{2}}{31}}$

$\boxed{\bf{\underline{\therefore \quad Rationalised \: Term = \dfrac{67 - 42 \sqrt{2}}{31}}}}$

Which is the required solution for this type of query.

Hope this helps you and clears the doubts for rationalising the terms!!!!!