If $\alpha , \beta , \gamma$ are roots of equation $x^3 + qx -r = 0$ then the equation, whose roots are

$\left( {\beta \gamma  + \frac{1}{\alpha }} \right),\,\left( {\gamma \alpha  + \frac{1}{\beta }} \right),\,\left( {\alpha \beta  + \frac{1}{\gamma }} \right)$

Suppose that $x$ and $y$ are positive number with $xy = \frac{1}{9};\,x\left( {y + 1} \right) = \frac{7}{9};\,y\left( {x + 1} \right) = \frac{5}{{18}}$ . The value of $(x + 1) (y + 1)$ equals

Let $f(x)=x^4+a x^3+b x^2+c$ be a polynomial with real coefficients such that $f(1)=-9$. Suppose that $i \sqrt{3}$ is a root of the equation $4 x^3+3 a x^2+2 b x=0$, where $i=\sqrt{-1}$. If $\alpha_1, \alpha_2, \alpha_3$, and $\alpha_4$ are all the roots of the equation $f(x)=0$, then $\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2$ is equal to. . . . . .

If $|x - 2| + |x - 3| = 7$, then $x =$

Let $p, q$ and $r$  be real numbers $(p \ne q,r \ne 0),$ such that the roots of the equation $\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to .

The solution set of the equation $pq{x^2} - {(p + q)^2}x + {(p + q)^2} = 0$ is