Question

If $x$ is real, the maximum values of expression $\frac{(x^2+14x+9)}{(x^2+2x+3)}$ will be?

Answer 1

Solve for maximum values

Let, $y=\frac{x^2+14x+9}{x^2+2x+3}$

$$\begin{gathered} \Rightarrow x^2+14x+9=(x^{2}y+2xy+3y) \\ \Rightarrow (1–y)x^2+(7-y)2x+3(3–y)=0 \end{gathered}$$

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

Since $x$ is real, therefore, $\sqrt{b^2–4ac}\ge 0$

$$\begin{gathered} \Rightarrow {\left[2\right(7–y\left)\right]}^2–4[1–y]\times 3[3–y]\ge 0 \\ \Rightarrow {\left(14-2y\right)}^2-12\left(1-y\right)\left(3-y\right)\ge 0 \\ \Rightarrow 196-56y+4y^2-12\left(3-y-3y+y^2\right)\ge 0 \\ \Rightarrow 196-56y+4y^2-36+48y-12y^2\ge 0 \\ \Rightarrow -8y^2-8y+160\ge 0 \\ \Rightarrow y^2+y-20\le 0 \\ \Rightarrow (y+5)(y–4)\le 0 \\ \Rightarrow -5\le y\le 4 \end{gathered}$$

Hence, the range in which the value of the given quadratic expression will have maximum value lie is $-5\le y\le 4$