Quadratic With Inequality Insane Level!

As $x$ only varies from $(-1,1)$ , so $x+6$ > 0

now if $k<0$ , then the situation is not possible which one can easily check himself/herself in two steps, So let us move on,

if $k>0$ , then we can simply move the denominator to the right and we get the quadratic

$x^2+k^2-kx-6\ge 0$

Now there are two ways for this and we must check both cases

$case(a)$ - the equation has roots but no root lies between [-1,1] and further that the minima point $\frac {-b}{2a}$ lies outside [-1,1}

since $k\ge 0$ so we automatically need $\frac {k}{2}\ge 1$ so $k\ge 2$

also we need $f(-1)f(1)\ge 0$

or $(k^2-7k+1)(k^2-5k+1)\ge 0$

we simply have to observe that this is positive for large k and as we approach along the number line from $\infty$ will first become negative at the largest root of the equation, which is the greater root of the 1st quadratic namely $\frac {7+3\sqrt{5}}{2}$

so $k\ge \frac {7+3\sqrt{5}}{2}$

which is answer

checking the other possibility where the equation has no root yields $k \ge 8$ which is clearly larger than $\frac {7+3\sqrt{5}}{2}$ and hence is not the answer

The problem can be solved by plotting $f(x,k) = \dfrac{x^2+k^2}{k(x+6)}$ for $x \in (-1,1)$ for various $k$ and inspect. It can be easily done with a spreadsheet.

The following shows the curves for integer $k$ from $-2$ to $8$ except $0$ .

It can be seen that the smallest $k$ , that is $a$ is slightly smaller than $7$ and that $f(x,a) = 1$ , that is the lowest point, when $x=1$ . Therefore,

$\dfrac {1^2+a^2}{a(1+6)} = 1\quad \Rightarrow a^2-7a + 1 = 0\quad \Rightarrow a = \boxed{6.854}$ (the other smaller root is unacceptable).

i solved it my third attempt :( , can you help what was wrong with my approach , i thought if the function i.e $x^2+k^2-kx-6k\ge 0$ is increasing , and moreover it has positive value for x=-1 , i can say it would be greater than equal to 0 for x=1 too .....and the given condition would be satisfied, so , i put x=1 and obtained the values for k , put x=-1 , obtained values of k and derivated it and put greater then zero , , the answer i got was for the equation in which i put x=1 , this was how i got the answer !