If m - 1-3 - -8 and n - 1-3 - -8- find if n2 is equal to 17 - 6 -m- then value of m is-

We rationalize the denominator, n= #160; 1 / 3+√( 8 ) #160;× 3 √( 8 ) #160;/ 3 √( 8 ) #160; n= #160; 3 √( 8 ) #160;/ 9 8 n= #160; 3 ...

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Integration Using Substitution

If m = 1/3 ...

Question

If $m = \frac{1}{3 - \sqrt{8}}$ and $n = \frac{1}{3 + \sqrt{8}}$ , find if $n^{2}$ is equal to $17 - 6 \sqrt{m}$ , then value of m is?

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Solution

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m = \frac{1}{3 - \sqrt{8}}

n = \frac{1}{3 + \sqrt{8}}

m^{2}

17 + 6 \sqrt{n}

n

n = \frac{1}{3 + \sqrt{8}}

n^{2}

17 - 6 \sqrt{m}

m

{log}_{8} m + {log}_{8} \frac{1}{6} = \frac{2}{3}

m

{(\frac{m}{n})}^{\frac{3}{8}} + {(\frac{n}{m})}^{\frac{3}{8}} = - 18

{(\frac{m}{n})}^{\frac{1}{8}} + {(\frac{n}{m})}^{\frac{1}{8}}

m = \frac{1}{3 - \sqrt{8}}

n = \frac{1}{3 + \sqrt{8}}

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