Irrational Equation

$\begin{aligned} x^2 - x - 10 \sqrt{x^2 -x - 11} & = 22 \\ \color{#3D99F6}{x^2 - x - 11} + 11 - 10 \sqrt{\color{#3D99F6}{x^2 - x - 11}} & = 22 \quad \quad \small \color{#3D99F6}{\text{Let } y^2 = x^2 - x - 11} \\ y^2 + 11 - 10y & = 22 \\ y^2 -10y - 11 & = 0 \\ (y+1)(y-11) & = 0 \\ \Rightarrow y & = \begin{cases} - 1 & \color{#D61F06}{\text{Rejected since } \sqrt{x^2 - x - 11} \ge 0} \\ 11 & \color{#3D99F6}{\text{Accepted}} \end{cases} \\ \Rightarrow x^2 - x - 11 & = 11^2 \\ x^2 - x - 132 & = 0 \end{aligned}$

By Vieta's formulas, the sum of roots of $x$ is $\boxed{1}$ .

Consider x^2 -x-11 = A , Substituting the A in the Given Equation we get :-

New Equation :- A^2 -10A + 11=0 The value of A would be { (10 + Sqrt(56))/2 , (10 - Sqrt(56))/2) } = Real Number Let's say 'N'.

Putting the Value of A we get :-

Now x^2 -x-11 = N => x^2 -x -(11+N) =0

The Summation of Roots is -b/a for a Equation ax^2 + bx + c = 0..

So the sum of the Roots of x is [ - ( -1) / 1] = 1 (Ans)

equaltion:

x^{2} -x-11 -10*\sqrt(x^{2} -x-11) + 5^{2}=36