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
at least two of $a_1, a_2, a_3$ are equal
$a_1=a_2=a_3$
$a_1, a_2, a _3$ are all distinct
$a_1, a_2, a _3$ are all positive and distinct
If $\alpha ,\beta,\gamma$ are the roots of equation $x^3 + 2x -5 = 0$ and if equation $x^3 + bx^2 + cx + d = 0$ has roots $2 \alpha + 1, 2 \beta + 1, 2 \gamma + 1$ , then value of $|b + c + d|$ is (where $b,c,d$ are coprime)-
Solution of the equation $\sqrt {x + 3 - 4\sqrt {x - 1} } + \sqrt {x + 8 - 6\sqrt {x - 1} } = 1$ is
The number of integral values of $m$ for which the quadratic expression, $(1 + 2m)x^2 -2(1+ 3m)x + 4(1 + m),$ $x\in R,$ is always positive, is
The sum of the roots of the equation, ${x^2}\, + \,\left| {2x - 3} \right|\, - \,4\, = \,0,$ is
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 .