Math, asked by vanshikhadas, 2 months ago

PLEASE SOLVE THIS QUESTION. WITH FULL PROCEDURE. NEED URGENTLY. PLEASE SOLVE IT FAST.
question 5 part 3​

Attachments:

Answers

Answered by senboni123456
3

Step-by-step explanation:

We have,

 \frac{7 +  \sqrt{5} }{7 -  \sqrt{5} } -  \frac{7 -  \sqrt{5} }{7 +  \sqrt{5} }  = a +  \frac{7}{11} b \sqrt{5}   \\

  \implies\frac{(7 +  \sqrt{5})^{2}   - (7 -  \sqrt{5}) ^{2}  }{(7 -  \sqrt{5} )(7 +  \sqrt{5} )} = a +  \frac{7}{11} b \sqrt{5}   \\

  \implies\frac{(7 +  \sqrt{5})^{2}   - (7 -  \sqrt{5}) ^{2}  }{(7 )^{2} - ( \sqrt{5} )^{2} } = a +  \frac{7}{11} b \sqrt{5}   \\

  \implies\frac{4.7.  \sqrt{5} }{49 - 5 } = a +  \frac{7}{11} b \sqrt{5}   \\

  \implies\frac{4.7.  \sqrt{5} }{44} = a +  \frac{7}{11} b \sqrt{5}   \\

  \implies\frac{7\sqrt{5} }{11} = a +  \frac{7}{11} b \sqrt{5}   \\

  \implies(0 )+ \frac{7}{11}(1) \sqrt{5}  = a +  \frac{7}{11} b \sqrt{5}   \\

On comparing both sides,

we get,  a=0 \: \& \: b=1

Similar questions