The locus of the point $P=(a, b)$ where $a, b$ are real numbers such that the roots of $x^3+a x^2+b x+a=0$ are in arithmetic progression is

Let $f: R \rightarrow R$ be the function $f(x)=\left(x-a_1\right)\left(x-a_2\right)$ $+\left(x-a_2\right)\left(x-a_3\right)+\left(x-a_3\right)\left(x-a_1\right)$ with $a_1, a_2, a_3 \in R$.Then, $f(x) \geq 0$ if and only if

Let $\alpha, \beta$ be roots of $x^2+\sqrt{2} x-8=0$. If $\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$, then $\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}$ is equal to ............

If for a posiive integer $n$ , the quadratic equation, $x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right) + .\;.\;.\; + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right) = 10n$ has two consecutive integral solutions, then $n$ is equal to:

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

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

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