Question

find the value of a and b in the following . 3+√7/3-√7=a+b√7

Answer 1

Hi friend
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Your answer
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$\frac{(3 + \sqrt{7} )}{(3 - \sqrt{7}) } = a \: + \: b \sqrt{7} \\ \\ first \: we \: need \: to \: rationalise \: the \: denominator. \\ \\ = > \frac{(3 + \sqrt{7}) }{(3 - \sqrt{7} )} \times \frac{(3 + \sqrt{7}) }{(3 + \sqrt{7} )} \\ \\ = > \frac{(3 + \sqrt{7} ) {}^{2} }{(3) {}^{2} - ( \sqrt{7} ) {}^{2} } \\ \\ = > \frac{9 + 6 \sqrt{7} + 7}{9 - 7} \\ \\ = > \frac{16 + 6 \sqrt{7} }{2} \\ \\ = > \frac{16}{2} + \frac{6 \sqrt{7} }{2} = a \: + \: b \sqrt{7} \\ \\ = > 8 + 3 \sqrt{7} = a \: + \: b \sqrt{7} \\ \\ therefore \\ \\ a \: = 8 \: and \: b \: = 3$

HOPE IT HELPS

Answer 2

Hi friend

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Your answer

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\begin{lgathered}\frac{(3 + \sqrt{7} )}{(3 - \sqrt{7}) } = a \: + \: b \sqrt{7} \\ \\ first \: we \: need \: to \: rationalise \: the \: denominator. \\ \\ = > \frac{(3 + \sqrt{7}) }{(3 - \sqrt{7} )} \times \frac{(3 + \sqrt{7}) }{(3 + \sqrt{7} )} \\ \\ = > \frac{(3 + \sqrt{7} ) {}^{2} }{(3) {}^{2} - ( \sqrt{7} ) {}^{2} } \\ \\ = > \frac{9 + 6 \sqrt{7} + 7}{9 - 7} \\ \\ = > \frac{16 + 6 \sqrt{7} }{2} \\ \\ = > \frac{16}{2} + \frac{6 \sqrt{7} }{2} = a \: + \: b \sqrt{7} \\ \\ = > 8 + 3 \sqrt{7} = a \: + \: b \sqrt{7} \\ \\ therefore \\ \\ a \: = 8 \: and \: b \: = 3\end{lgathered}

(3−

7

)

(3+

7

)

=a+b

7

firstweneedtorationalisethedenominator.

=>

(3−

7

)

(3+7) × (3+7 )

(3+7 )

=>

(3) 2−(72)

(3+ 72)

=>

9−7

9+6

7+7

=>

2

16+6

7

=>

2

16+ 26 7

=a+b

7

=>8+3

7

=a+b

7

Therefore

a=8andb=3