Graph of $\mathrm{y}={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$ is as shown in the figure. If PQ = 9, OR = 5 and OB = 2.5, then which of the following is/are true?

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{OR}=5\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{f}\left(0\right)=-5=\mathrm{c}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{PQ}=9\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-\frac{\mathrm{D}}{4\mathrm{a}}=-9\end{array}$

$\begin{array}{l}\Rightarrow \frac{{\mathrm{b}}^2+20\mathrm{a}}{4\mathrm{a}}=9\\ \Rightarrow {\mathrm{b}}^2=16\mathrm{a} ...... \left(1\right)\\ \mathrm{OB}=2.5\end{array}$

So, one root of equation is 2.5.

$\begin{array}{l}\therefore 25\mathrm{a}+10\mathrm{b}-20=0\\ \text{ or }5\mathrm{a}+2\mathrm{b}-4=0 ........ \left(2\right)\end{array}$

From (1) and (2),

$\begin{array}{r}5{\mathrm{b}}^2+32\mathrm{b}-64=0\\ \Rightarrow (5\mathrm{b}-8)(\mathrm{b}+8)=0\end{array}$

$\Rightarrow \mathrm{b}=-8,\frac{8}{5}$ (Not possible. Since a> 0, we must have b < 0)

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{a}=4\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{y}=4{\mathrm{x}}^2-8\mathrm{x}-5=4{\mathrm{x}}^2-10\mathrm{x}+2\mathrm{x}-5=(2\mathrm{x}-5)(2\mathrm{x}+1)\end{array}$

So, x = 2.5, -0.5 are the roots of y(x) = 0.

Clearly, y(-1) > 0.

$\begin{array}{l}\mathrm{y}\ge 7\\ \Rightarrow 4{\mathrm{x}}^2-8\mathrm{x}-5\ge 7\\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}-3\ge 0\\ \Rightarrow (\mathrm{x}-3)(\mathrm{x}+1)\ge 0\\ \Rightarrow \mathrm{x}\le -1\text{ or }\mathrm{x}\ge 3\\ {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=\mathrm{mx}\\ \Rightarrow 4{\mathrm{x}}^2-(8+\mathrm{m})\mathrm{x}-5=0\end{array}$

Clearly, above equation has two distinct real roots for any real value of m.