Question

rationalize the denominator 3 by under root 7 minus under root 2

Answer 1

Rationalizing the denominator :

$\tt {\frac{3}{\sqrt{7} - \sqrt{2}} }$

Step by step explanation :

Multiply the fraction ( numerator and denominator ) by the conjugate of the denominator.

$\tt {\therefore \frac{3}{\sqrt{7} - \sqrt{2} } \times \frac{\sqrt{7} + \sqrt{2} }{\sqrt{7} + \sqrt{2}} }$

$\tt{ \frac{3(\sqrt{7} + \sqrt{2} )}{(\sqrt{7} - \sqrt{2}) (\sqrt{7} + \sqrt{2})} }$

Here, In the denominator, Simplify it by using the identity :

( a + b ) ( a - b ) = a^2 - b^2

$\tt {\therefore \frac{3 \sqrt{ 7}+ 3 \sqrt{2} }{7- 2}}$

$\tt {= \frac{3(\sqrt{7} + \sqrt{2} )}{47}}$

Thus, The rationalised denominator form is :

$\tt{\frac{ 3(\sqrt 7+ \sqrt{2}) }{47}}$

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Answer 2

HOPE IT HELPS YOU !!

3/√7-√2 = 3/√7-√2  ×√7+√2/√√7+2

              = 3(√7+√2)/ (√7)² -(√2)²                 {(a+b)(a-b)= a²-b²}

              =3(√7+√2)/49-2

              = 3(√7+√2)/47

              =3(√7+√2) / 47