The equation $\sqrt {3 {x^2} + x + 5} = x - 3$ , where $x$ is real, has
no solution
exactly one solution
exactly two solution
exactly four solution
If $\alpha , \beta , \gamma $ are roots of equation ${x^3} + a{x^2} + bx + c = 0$, then ${\alpha ^{ - 1}} + {\beta ^{ - 1}} + {\gamma ^{ - 1}} = $
The solutions of the quadratic equation ${(3|x| - 3)^2} = |x| + 7$ which belongs to the domain of definition of the function $y = \sqrt {x(x - 3)} $ are given by
The number of real roots of the equation $\mathrm{e}^{4 \mathrm{x}}-\mathrm{e}^{3 \mathrm{x}}-4 \mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{\mathrm{x}}+1=0$ is equal to $.....$
If $\alpha , \beta , \gamma$ are roots of equation $x^3 + qx -r = 0$ then the equation, whose roots are
$\left( {\beta \gamma + \frac{1}{\alpha }} \right),\,\left( {\gamma \alpha + \frac{1}{\beta }} \right),\,\left( {\alpha \beta + \frac{1}{\gamma }} \right)$
Let $t$ be real number such that $t^2=a t+b$ for some positive integers $a$ and $b$. Then, for any choice of positive integers $a$ and $b, t^3$ is never equal to