Algebra and Polynomials

${ x }^{ 2 }-1=5x\\ \Rightarrow { x }-\frac { 1 }{ { x } } =5\\ \Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ \left( { x }-\frac { 1 }{ { x } } \right) }^{ 2 }+2 \\ ={ 5 }^{ 2 }+2 = \boxed{27}$

here it is:

       x^2-1=5x

         or, x-1/x=5................................[dividing the both sides  by x]


          or,(x-1/x)^2=5^2


       or,(x + 1/x)^2  -4 .x . 1/x=25....................................[(a-b)^2=(a+-b)^2- 4 . a .b)]


        or, (x + 1/x)^2 =29


         or, x^2+ 1/x^2 + 2 . x .1/x=29


            or, x^2+ 1/x^2=29-2=27

Solution using the quadratic formula:

$x^2-5x-1=0$ $\implies$ $a=1,b=-5$ and $c=-1$

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

$=\dfrac{5 \pm \sqrt{25-4(1)(-1)}}{2}$

$=\dfrac{5 \pm \sqrt{29}}{2}$

Using the positive value of $x$ , we get

$x^2+\dfrac{1}{x^2} = \left(\dfrac{5+\sqrt{29}}{2} \right)^2+\dfrac{1}{\left(\dfrac{5+\sqrt{29}}{2} \right)^2}=$ $\color{#D61F06}\large \boxed{27}$

I completed the square to get x = 5/2 + √(29/4) When I put that back into the equation I got 27