Let $\alpha, \beta$ be two roots of the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$. Then $\alpha^{8}+\beta^{8}$ is equal to:

Let $p, q$ be integers and let $\alpha, \beta$ be the roots of the equation, $x^2-x-1=0$, where $\alpha \neq \beta$. For $n=0,1,2, \ldots$, let $a_n=$ $p \alpha^n+q \beta^n$.

$FACT$ : If $a$ and $b$ are rational numbers and $a+b \sqrt{5}=0$, then $a=0=b$.

($1$) $a_{12}=$

$[A]$ $a_{11}-a_{10}$  $[B]$ $a_{11}+a_{10}$  $[C]$ $2 a_{11}+a_{10}$   $[D]$ $a_{11}+2 a_{10}$

($2$) If $a_4=28$, then $p+2 q=$

$[A] 21$   $[B] 14$   $[C] 7$    $[D] 12$

 answer the quetion ($1$) and ($2$)

If graph of $y = ax^2 -bx + c$ is following, then sign of $a$, $b$, $c$ are

The equation $x^2-4 x+[x]+3=x[x]$, where $[x]$ denotes the greatest integer function, has:

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:

If $\alpha$ and $\beta$ are the distinct roots of the equation $x^{2}+(3)^{1 / 4} x+3^{1 / 2}=0$, then the value of $\alpha^{96}\left(\alpha^{12}-\right.1) +\beta^{96}\left(\beta^{12}-1\right)$ is equal to: