Question

If x^(4)+x^(2)=(11)/(5) then find value of root(3)((x+1)/(x-1))+root(3)((x-1)/(x+1))

Answer 1

Answer:

f the equations x^ 2 -px=0\&x^ 2 +px-p=0 have a common root, then the reciprocal of sum of all possible value(s) of p is ​

Step-by-step explanation:

If x^(4)+x^(2)=(11)/(5) then find value of root(3)((x+1)/(x-1))+root(3)((x-1)/(x+1))

Answer 2

Given:

$x^{4} + x^{2} = \frac{11}{5}$

To find:

The value of  $\sqrt{3}\frac{x+1}{x-1}+\sqrt{3}\frac{x-1}{x+1}$

Solution:

We have,

$x^{4} + x^{2} = \frac{11}{5}$

If we add $\frac{1}{4}$ both sides, we get

$x^{4} + x^{2} +\frac{1}{4}= \frac{11}{5}+\frac{1}{4}$

Using the identity , $(a+b)^{2} = a^{2} + 2ab+b^{2}$ , we get

$(x^{2} + \frac{1}{2} )^{2} =\frac{49}{20} ; or\\x^{2} + \frac{1}{2}= \frac{7}{2\sqrt{5} } ; or\\x^{2} = \frac{7}{2\sqrt{5} } - \frac{1}{2}$

Hence,

$x^{2} = \frac{7-\sqrt{5} }{2\sqrt{5} }$    (i)

Now,

We have ,

$=\sqrt{3}\frac{x+1}{x-1}+\sqrt{3}\frac{x-1}{x+1}$

Solving this we get,

$=\sqrt{3}(\frac{x+1}{x-1}+\frac{x-1}{x+1} ) \\=\sqrt{3}(\frac{(x+1)^{2}+(x-1)^{2}}{(x-1)(x+1)} )$

Applying the identities, we get

$=\sqrt{3}\frac{(x^{2} +2x+1+x^{2} -2x+1)}{x^{2} -1}$ ; or

$=\sqrt{3}(\frac{2(x^{2} +1)}{x^{2} -1} )$

Applying the value of $x^{2}$ calculated in (i) in the above equation, we get

$=2\sqrt{3}(\frac{(\frac{7-\sqrt{5} }{2\sqrt{5} } )+1}{(\frac{7-\sqrt{5} }{2\sqrt{5} } )-1} )$  ; or

$=2\sqrt{3}(\frac{7-\sqrt{5}+2\sqrt{5} }{7-\sqrt{5}-2\sqrt{5} } )$ ; and hence,

$=2\sqrt{3}(\frac{7+\sqrt{5} }{7-3\sqrt{5} } )$

We can simplify it further,

$= 2\sqrt{3}(\frac{7+\sqrt{5} }{7-3\sqrt{5} } \frac{(7+3\sqrt{5}) }{(7+3\sqrt{5}) } ) \\= 2\sqrt{3}\frac{(49+21\sqrt{5}+7\sqrt{5}+15 )}{(49-45)} \\= 2\sqrt{3}(\frac{64+28\sqrt{5} }{4} ) \\=2\sqrt{3}(16+7\sqrt{5} ) \\= 32\sqrt{3}+ 14\sqrt{15}$

Final answer:

Hence the value of $\sqrt{3}\frac{(x+1)}{(x-1)} +\sqrt{3}\frac{(x-1)}{(x+1)}$ is $32\sqrt{3}+ 14\sqrt{15}$.