Math, asked by abhdk2, 9 months ago

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

Answers

Answered by crystalch24092
16

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...

Answered by ItzDevilQueen07
7

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...

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