Question

rationalize the denominator;;1/7+3 root 2

Answer 1

Question is to Rationalize : $\frac{1}{7+3\sqrt{2} }$

Solution :

  $\frac{1}{7+3\sqrt{2} }$

= $\frac{1}{7+3\sqrt{2} }( \frac{7-3\sqrt{2}}{7-3\sqrt{2} })$               [ Multiplying by $\frac{7-3\sqrt{2}}{7-3\sqrt{2} }$ ]

=  $\frac{7-3\sqrt{2}}{(7)^{2} -(3\sqrt{2})^{2} }$

=$\frac{7-3\sqrt{2}}{49 -18 }$

=$\frac{7-3\sqrt{2}}{31}$

Hence, the denominator is rationalized.

                                                                                 

Hope this helped you...

Answer 2

Answer:

Question is to Rationalize : \frac{1}{7+3\sqrt{2} }

7+3

2

1

Solution :

\frac{1}{7+3\sqrt{2} }

7+3

2

1

= \frac{1}{7+3\sqrt{2} }( \frac{7-3\sqrt{2}}{7-3\sqrt{2} })

7+3

2

1

(

7−3

2

7−3

2

) [ Multiplying by \frac{7-3\sqrt{2}}{7-3\sqrt{2} }

7−3

2

7−3

2

]

= \frac{7-3\sqrt{2}}{(7)^{2} -(3\sqrt{2})^{2} }

(7)

2

−(3

2

)

2

7−3

2

=\frac{7-3\sqrt{2}}{49 -18 }

49−18

7−3

2

=\frac{7-3\sqrt{2}}{31}

31

7−3

2

Hence, the denominator is rationalized.

Hope this helped you...