good question on quadratic i guess.....

well 1 more...!!

if (y² - 5y + 3)(x² + x + 1) < 2x for all x E R, then find the interval in which y lies.

For second one, I'm seem not to get a very ' neat ' answer. Would you mind telling the answer.

for the second one.....

5 - √52 < y < 5 + √52

Yes, I got it then.

See, you just send the ' x ' term on left side on right side and then find the minimum value of RHS.
This will come out as -2. So you have,

y²-5y+3<-2

Now Solve to obtain the above condition for y.

\hspace{-16}\mathbf{\left(x^2-5y+3\right).\left(x^2+x+1\right)<2x\Leftrightarrow \left(y^2-5y+3\right)<\frac{2x}{x^2+x+1}}\\\\ $Now first we will Calculate $\mathbf{Min.}$ of $\mathbf{\frac{2x}{x^2+x+1}}$\\\\ For This, Let $\mathbf{z=\frac{2x}{x^2+x+1}\Leftrightarrow zx^2+zx+z=2x}$\\\\ $\mathbf{zx^2+x.(z-2)+z=0}$\\\\ Now If Given Equation has $\mathbf{Real}$ Roots\;,Then its $\mathbf{D\geq 0}$\\\\ so $\mathbf{(z-2)^2-4.z^2\geq 0\Leftrightarrow (2z)^2-(z-2)^2\leq 0}$\\\\ $\mathbf{(3z-2).(z+2)\leq 0\Leftrightarrow 3.\left(\frac{2}{3}\right).(z+2)\leq 0}$\\\\ So $\mathbf{-2\leq z\leq \frac{2}{3}}$\\\\ So $\mathbf{\left(y^2-5y+3\right)<\frac{2x}{x^2+x+1}\leq -2}$\\\\ So $\mathbf{y^2-5y+5<0}$\\\\ So $\mathbf{\left(y-\left(\frac{5-\sqrt{5}}{2}\right)\right).\left(y-\left(\frac{5+\sqrt{5}}{2}\right)\right)<0}$\\\\\\ So $\mathbf{\frac{5-\sqrt{5}}{2}<y<\frac{5+\sqrt{5}}{2}}$\\\\\\ $*\;$ Max. and Min can also be obtained Using $\mathbf{A.M\geq G.M}$\\\\

\hspace{-16}(1)\;\;$Let $\mathbf{y=\frac{x^2+ax+3}{x^2+x+a}\Leftrightarrow yx^2+xy+ay=x^2+ax+3}$\\\\\\ $\mathbf{(y-1).x^2+x.(y-a)+(ay-3)=0}$\\\\ Now Given equation has Real Roots\\\\ So its $\mathbf{D\geq 0}$\\\\ $\mathbf{(y-a)^2-4.(y-1).(ay-3)\geq 0}$\\\\\\ $\mathbf{(1-4a).y^2+2y(a+6)+(a^2-12)\geq 0}$\\\\\\ Now If $\mathbf{(1-4a)>0\;}$ Then It Represent $\mathbf{Upward}$ Parabola\\\\ So its $\mathbf{D\leq 0}$\\\\ $\mathbf{4.(a+6)^2-4.(1-4a).(a^2-12)\leq 0}$\\\\\\ $\mathbf{4a^3+48-36a\leq 0}$\\\\ But Here $\mathbf{(1-4a)>0\Leftrightarrow 4a<1\Leftrightarrow 36a<9}$\\\\\\ So $\mathbf{4a^3+48-36a\leq 0}$\\\\ So $\mathbf{4a^3+39<0}$\\\\

wonderfl sir.. :P