If $x$ is real,then the value of $x^2 - 6x + 13$ will not be less than

The quadratic equation $x^2+15|x|+14=0$ has

Let $a, b, c$ be three real numbers such that $a + 2b + 4c = 0$. Then the equation $ax^2 + bx + c = 0$

The values of $a$ for which the expression $(a^2 - 1)x^2 + 2(a - 1)x + 2$ is positive for all $x \in \mathbb{R}$ are:

If $\alpha, \beta, \gamma, \delta, \varepsilon$ are the roots of the equation $x^5+x^4-13x^3-13x^2+36x+36=0$ and $\alpha < \beta < \gamma < \delta < \varepsilon$,then find the value of $\frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}$.

Let $x_1, x_2, x_3 \in \mathbb{R} \setminus \{0\}$,$x_1 + x_2 + x_3 \neq 0$ and $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$. Then $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$ holds good for: