Question

If x=c√b+4 then find the value of x+1/x

Answer 1

x+ 1/x = c√b + 4 + 1 / (c√b + 4)

= ((c√b+4)^2 +1) /(c√b+4)

= (c^2 b + 8 c√b + 16+1) / (c√b+4)

= (c^2 b + 8 c√b + 17) / (c√b+4)

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

$\rm{x = c \sqrt{5} + 4}$

$\bf \underline{To \: find-}$

$\rm{the \: value \: of \: x + \frac{1}{x} =? }$

$\bf \underline{Solution-}$

$\textsf{We have}$

$\rm{x = c \sqrt{b} + 4}$

$\rm{ \therefore \: \frac{1}{x} = \frac{1}{c \sqrt{b} + 4} }$

$\textsf{Here the denominator is c√b + 4}$

$\textsf{We know that,}$

$\textsf{Rationalising factor of a√b+c = a√b-c}$

$\textsf{On rationalising the denominator them}$

$\rm{ \frac{1}{x} = \frac{1}{c \sqrt{b} + 4} \times \frac{c \sqrt{b} - 4 }{c \sqrt{b} - 4} }$

$\rm{ \frac{1}{x} = \frac{1(c \sqrt{b} - 4)}{(c \sqrt{b} + 4)(c \sqrt{b} - 4) } }$

$\rm{ \frac{1}{x} = \frac{c \sqrt{b} - 4}{(c \sqrt{b} + 4)(c \sqrt{b} - 4) } }$

$\textsf{Comparing the given expression in denominator with }$

$\textsf{(a+b) (a - b), we get}$

$\textsf{a = c√b and b = 4.}$

$\textsf{Now using, (a + b)(a - b²) = a² - b², we get}$

$\rm{ \frac{1}{x} = \frac{c \sqrt{b} - 4}{(c \sqrt{b} {)}^{2} - (4 {)}^{2} } }$

$\rm{ \frac{1}{x} = \frac{c \sqrt{b} - 4}{b {c}^{2} - 16 } }$

$\textsf{Now,adding both values x and 1/x}$

$\rm{ \therefore \: x + \frac{1}{ x} = c \sqrt{b} + 4 + \frac{c \sqrt{b} - 4 }{b {c}^{2} - 16} }$

$\rm{ x + \frac{1}{ x} = \frac{(c \sqrt{b} + 4)(b {c}^{2} - 16)+c \sqrt{b} - 4 }{b {c}^{2} - 16 } }$

$\rm{ x + \frac{1}{ x} = \frac{ {c}^{3}b \sqrt{b} - 15c \sqrt{b} + 4b {c}^{2} - 68 }{b {c}^{2} - 16 } }$

$\rm{ \underline{Hence, \: the \: value \: of \: x + \frac{1}{ x} \: is \: \frac{ {c}^{3}b \sqrt{b} - 15c \sqrt{b} + 4b {c}^{2} - 68 }{b {c}^{2} - 16 } .}}$