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  • If sqroot 2 =1.414 and sqroot 3 =1.732 then find the value of 4 by 3 sqroot 3-2sqroot2 +3 by3sqroot3+2sqroot 2

If \sqrt{2}= 1\cdot 414 and \sqrt{3}= 1\cdot 732 then find the value of \frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}

Answers (1)

Answer.     2.0632
Solution.   Given that :
\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732
\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}
= \frac{4\left ( 3\sqrt{3}+2\sqrt{2} \right )}{\left ( 3\sqrt{3}-2\sqrt{2} \right )\left ( 3\sqrt{3}+2\sqrt{2} \right )}+\frac{3\left ( 3\sqrt{3}-2\sqrt{2} \right )}{\left ( 3\sqrt{3}+2\sqrt{2} \right )\left (3\sqrt{3}-2\sqrt{2} \right )}
= \frac{4\left ( 3\sqrt{3}+2\sqrt{2} \right )+3\left ( 3\sqrt{3}-2\sqrt{2} \right )}{\left ( 3\sqrt{3}-2\sqrt{2} \right )\left ( 3\sqrt{3}+2\sqrt{2} \right )}
Using   (a – b) (a + b) = a2 – b2
= \frac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{\left ( 3\sqrt{3} \right )^{2}-\left ( 2\sqrt{2} \right )^{2}}
= \frac{21\sqrt{3}+2\sqrt{2}}{27-8}
= \frac{21\sqrt{3}+2\sqrt{2}}{19}
Putting the given values,= \frac{21\left ( 1\cdot 732 \right )+2\left ( 1\cdot 414 \right )}{19}
=\frac{39\cdot 2014}{19}
= 2.0632
Hence the answer is 2.0632.

Posted by

infoexpert27

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