Question

Find the roots of the equation x+1/x=3 ,​

Answer 1

Solution:

$\tt{x + \dfrac{1}{x} = 3}$

Solve the LHS by taking LCM

$\dashrightarrow \: \: \tt{ \dfrac{ {x}^{2} + 1}{x} = 3}$

Transpose x to RHS

$\dashrightarrow \: \: \tt{ {x}^{2} + 1 = 3x}$

Bring all terms to LHS so that it would be easier to factorise.

$\dashrightarrow \: \: \tt{ {x}^{2} - 3x + 1 = 0}$

Factorise the polynomial by applying quadratic formula

$\dashrightarrow \: \: \tt{ x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}}$

Now,

$\dashrightarrow \: \: \tt{x = \dfrac{ - ( - 3) \pm \sqrt{( - 3)^{2} - 4(1)(1) } }{2(1)} }$

On further simplifying

$\dashrightarrow \: \: \tt{x = \dfrac{3 \pm \sqrt{9 - 4} }{2} }$

So,

$\boxed{ \bf{x = \dfrac{3 \pm \sqrt{5} }{2}}}$

Answer 2

Step-by-step explanation:

$\\ \sf{:}\longrightarrow x+\dfrac{1}{x}=3$

$\\ \sf{:}\longrightarrow \dfrac{x^2+1}{x}=3$

$\\ \sf{:}\longrightarrow x^2+1=3x$

$\\ \sf{:}\longrightarrow x^2-3x+1=0$

$\\ \sf{:}\longrightarrow x^2-3x=-1$

$\\ \sf{:}\longrightarrow 4(x^2-3x)=4(-1)$

$\\ \sf{:}\longrightarrow 4x^2-12x=-4$

$\\ \sf{:}\longrightarrow (2x)^2-2(2x)(3)=-4$

$\\ \sf{:}\longrightarrow (2x)^2-2(2x)(3)+(3)^2=(3)^2-4$

$\\ \sf{:}\longrightarrow (2x-3)^2=9-4$

$\\ \sf{:}\longrightarrow (2x-3)^2=5$

$\\ \sf{:}\longrightarrow \sqrt{(2x-3)^2}=\sqrt{5}$

$\\ \sf{:}\longrightarrow 2x-3=\underline{+}\sqrt{5}$

$\\ \sf{:}\longrightarrow 2x=3\underline{+}\sqrt{5}$

$\\ \sf{:}\longrightarrow{\underline{\boxed{\bf x=\dfrac{3\underline{+}\sqrt{5}}{2}}}}$