Question

if m^4+1/m^4=194,find m^3+1/m^3,m^2+1/m^2and m+1/m.

Answer 1

Here's your answer..

Solution:-
Given :-
$= > {m}^{4} + \frac{1}{ {m}^{4} } = 194 \\ = > {( {m}^{2} + \frac{1}{ {m}^{2} } ) }^{2} - 2 = 194$
$= > {( {m}^{2} + \frac{1}{ {m}^{2} }) }^{2} = 194 + 2 \\ = > {( {m }^{2} + \frac{1}{ {m}^{2} }) }^{2} = 196$
$= > {( {m}^{2} + \frac{1}{ {m}^{2} } ) }^{2} = \sqrt{196}$

$= > {( {m}^{2} + \frac{1}{ {m}^{2} }) }^{2} = 14.......(1)$

Now,
$= > {(m + \frac{1}{m} )}^{2} - 2 = 14 \\ = > {(m + \frac{1}{m} )}^{2} = 14 + 2$

$= > {(m + \frac{1}{m} )}^{2} = 16$
$= > {(m + \frac{1}{m} ) }^{2} = \sqrt{16 } \\ = > (m + \frac{1}{m} ) = 4........(2)$

We know that,

$= > {m}^{3} + \frac{1}{ {m}^{3} } = (m + \frac{1}{m} )( {m}^{2} + \frac{1}{ {m}^{2} } - 1)$

From 2 and 3 ,we get:-

$= > {m}^{3} + \frac{1}{ {m}^{3} } = 4(14 - 1) \\ = > {m}^{3} + \frac{1}{ {m}^{2} } = 4 \times 13 \\ = > {m}^{3} + \frac{1}{ {m}^{3} } = 52$

Therefore,

$(a) = > {m}^{3} + \frac{1}{ {m}^{3} } = 52$

From (2)

$(b) = > m + \frac{1}{m} = 4$


Hope it helps you...