Math, asked by santoshsubedi992, 1 year ago

If (root3 +2 root 5)/(root3 -2 root 5)=(B + A root15) then value of A and B are?

Answers

Answered by TPS
1
 \frac{ \sqrt{3}+2 \sqrt{5}  }{ \sqrt{3}-2 \sqrt{5} } =B+A \sqrt{15} \\ \\ \Rightarrow  \frac{ \sqrt{3}+2 \sqrt{5}  }{ \sqrt{3}-2 \sqrt{5} }\ \times\ \frac{ \sqrt{3}+2 \sqrt{5}  }{ \sqrt{3}+2 \sqrt{5} } =B+A \sqrt{15} \\ \\ \Rightarrow \frac{ (\sqrt{3}+2 \sqrt{5}) \times(\sqrt{3}+2 \sqrt{5} )}{ (\sqrt{3}-2 \sqrt{5}) \times(\sqrt{3}+2 \sqrt{5} )}=B+A \sqrt{15}

\Rightarrow  \frac{ (\sqrt{3})^2+(2 \sqrt{5})^2+2 \times \sqrt{3} \times 2 \sqrt{5}}{ (\sqrt{3})^2-(2 \sqrt{5})^2}=B+A \sqrt{15} \\ \\ \Rightarrow  \frac{3+20+4 \sqrt{15} }{3-20} =B+A \sqrt{15} \\ \\ \Rightarrow  \frac{23+4 \sqrt{15} }{-17} =B+A \sqrt{15} \\ \\ \Rightarrow -\frac{23}{17}  - \frac{4}{17}\sqrt{15} =B+A \sqrt{15}\\ \\ Thus\ A=-\frac{4}{17},\ B=-\frac{23}{17}

santoshsubedi992: thank you
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