The product of all solutions of the equation $e^{5(\ln x)^2+3} = x^8$,where $x > 0$,is:

If $m, n$ are the roots of the equation ${x^2} - x - 1 = 0$,then the value of $\frac{{\left( {1 + m{{\log }_e}3 + \frac{{{{(m{{\log }_e}3)}^2}}}{{2!}} + ...\infty } \right)\left( {1 + n{{\log }_e}3 + \frac{{{{(n{{\log }_e}3)}^2}}}{{2!}} + ...\infty } \right)}}{{\left( {1 + mn{{\log }_e}3 + \frac{{{{(mn{{\log }_e}3)}^2}}}{{2!}} + ...\infty } \right)}}$ is:

The number of values of $b$ for which there is an isosceles triangle with sides of lengths $b+5$,$3b-2$,and $6-b$ is

Let $a, b, c$ be real numbers with $a \ne 0$. If $\alpha$ is a root of $a^2x^2 + bx + c = 0$,$\beta$ is a root of $a^2x^2 - bx - c = 0$,and $0 < \alpha < \beta$,then the equation $a^2x^2 + 2bx + 2c = 0$ has a root $\gamma$ that always satisfies:

If $(x-1)$ is a factor of the polynomial $x^{5}-4 x^{3}+2 x^{2}-3 x+k=0$,then the value of $k$ is:

If the roots of the given equation $(\cos p-1) x^2+(\cos p) x+\sin p=0$ are real,then