Question

Rationalize the denominator 1 by 5 + 3 / 2

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

What is Rationalising??

Rationalising means removing root values from the denominator and shift towards numerator.

Note : There will always a identity used in the denominator in most cases =>(x+y)(x-y)

Method of Rationalising

step 1: Firstly multiply both numerator and denominator with the opposite sign of denominator.

Step2: Then a identity will use in the denominator and hence we are able to remove root from the denominator and a single value left in the denominator.

$= > \frac{1}{5 + 3 \sqrt{2} }$

$= > \frac{1}{5 + 3 \sqrt{2} } \times \frac{5 - 3 \sqrt{2} }{5 - 3 \sqrt{2} }$

Here,this identity is used:

$\bold{\boxed{\purple{(x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

$= > \frac{5 - 3 \sqrt{2} }{ {(5)}^{2} - {(3 \sqrt{2} )}^{2} }$

$= > \frac{5 - 3 \sqrt{2} }{25 - 18}$

$= > \frac{5 - 3 \sqrt{2} }{7}$

$\bold\fcolorbox{aqua}{orange}{Answer}$$\bold{\frac{5 - 3 \sqrt{2} }{7}}$

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

What is Rationalising??

Rationalising means removing root values from the denominator and shift towards numerator.

Note : There will always a identity used in the denominator in most cases =>(x+y)(x-y)

Method of Rationalising

step 1: Firstly multiply both numerator and denominator with the opposite sign of denominator.

Step2: Then a identity will use in the denominator and hence we are able to remove root from the denominator and a single value left in the denominator.

$= > \frac{1}{5 + 3 \sqrt{2} }$

$= > \frac{1}{5 + 3 \sqrt{2} } \times \frac{5 - 3 \sqrt{2} }{5 - 3 \sqrt{2} }$

Here,this identity is used:

$\bold{\boxed{\purple{(x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

$= > \frac{5 - 3 \sqrt{2} }{ {(5)}^{2} - {(3 \sqrt{2} )}^{2} }$

$= > \frac{5 - 3 \sqrt{2} }{25 - 18}$

$= > \frac{5 - 3 \sqrt{2} }{7}$

$\bold\fcolorbox{aqua}{orange}{Answer}$$\bold{\frac{5 - 3 \sqrt{2} }{7}}$