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**Hint:**Here in this question it is given that $x$ is real and also given an equation in $x$ and you have to prove that range of $x$ is between $ - \dfrac{1}{{11}}$ and $1$. In order to do so, you have to make a quadratic equation by considering it equal to another variable and then apply the discriminant conditions for having real roots.

**Complete step by step solution:**It is the question that is given an equation in $x$.

Let us consider $\dfrac{x}{{{x^2} - 5x + 9}} = y$

Now on cross multiplication, we get

$ \Rightarrow x = y{x^2} - (5y)x + 9y$

On making the above equation a quadratic in terms of $x$, we get

$y{x^2} - (1 + 5y)x + 9y = 0$ ……….(i)

Here in the question, it is also given that the $x$ is real.

Now according to the question, this quadratic equation must have real roots. In doing so, the discriminant for this quadratic equation must be greater than or equal to zero.

Discriminant must be $ \geqslant 0$

$\therefore {b^2} - 4ac \geqslant 0$ ……….(ii)

In order to find $a,b{\text{ and }}c$, compare equation (i) with the standard quadratic equation $a{x^2} + bx + c = 0$

On comparing, we get

$a = y$ , $b = - (1 + 5y)$ and $c = 9y$

Now putting all that values in equation (ii), we get

${( - 1 - 5y)^2} - 4 \times y \times 9y \geqslant 0$

$ \Rightarrow 1 + 25{y^2} + 10y - 36{y^2} \geqslant 0$

On further solving, we get

$ \Rightarrow 1 + 10y - 11{y^2} \geqslant 0$

Now, solving this equation using factorisation method

$ \Rightarrow 1 - y + 11y - 11{y^2} \geqslant 0$

On further making factors, we get

$ \Rightarrow (1 - y)(11{y^2} + 1) \geqslant 0$

Now on taking minus sign common, the inequality symbol gets reversed

$ \Rightarrow (11y + 1)(y - 1) \leqslant 0$

On solving this, we get range as

$y \in \left[ { - \dfrac{1}{{11}},1} \right]$

**Hence, it is proved that the $\dfrac{x}{{{x^2} - 5x + 9}}$ lies between $ - \dfrac{1}{{11}}$ and $1$**

**Note:**

The discriminant is the part of the quadratic formula underneath the square root symbol ${b^2} - 4ac$ . The discriminant tells us whether there are two solutions, one solution, or no solutions.

The discriminant can be positive, zero, or negative, and this determines how many solutions there are to the given quadratic equation.

-A positive discriminant indicates that the quadratic has two distinct real number solutions.

-A discriminant of zero indicates that the quadratic has a repeated real number solution.

-A negative discriminant indicates that neither of the solutions is real numbers.

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