Q.If $x$ is real, prove that $\dfrac{x}{x^2 - 5x + 9}$ lies between $-\dfrac{1}{11}$ and $1$.

Setting $y=\dfrac{x}{x^2-5x+9}$ and requiring the resulting quadratic in $x$ to have real roots (discriminant $\ge0$) shows $y$ is confined to $[-\tfrac{1}{11},1]$.

Step 1 — Check the denominator never vanishes.

$x^2-5x+9$ has discriminant $25-36=-11<0$, and the leading coefficient is positive, so $x^2-5x+9>0$ for every real $x$ — the expression is defined for all real $x$.

Step 2 — Let $y$ be a value taken by the expression.

$y=\dfrac{x}{x^2-5x+9} \implies y(x^2-5x+9)=x \implies yx^2-(5y+1)x+9y=0$.

This is a quadratic in $x$ (for $y\ne0$). For a real $x$ to exist, its discriminant must be $\ge0$.

Step 3 — Compute the discriminant.

$D=(5y+1)^2-4(y)(9y) = 25y^2+10y+1-36y^2 = -11y^2+10y+1 \ge 0$.

Multiply by $-1$ (flip inequality): $11y^2-10y-1\le0$.

Step 4 — Solve the quadratic inequality.

Roots of $11y^2-10y-1=0$: $y=\dfrac{10\pm\sqrt{100+44}}{22}=\dfrac{10\pm12}{22}$, giving $y=1$ or $y=-\dfrac{1}{11}$.

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