Question

Total number of values of ${}^{\prime }{a}^{\prime }$, so that $x^2-x-a=0$ has integral roots, where $aϵN$ and $6\le a\le 100$, is equal to

• A. $2$
• B. $4$
• C. $8$
• D. $6$

Answer 1

Answer: (C)

$8$

$x^2-x-a=0,D=1+4a=odd$
$D$ must be perfect square of some odd integer.
Let $D=(2\lambda +1{)}^2\Rightarrow 1+4a=1+4{\lambda }^2+4\lambda$
$\Rightarrow a=\lambda (\lambda +1)$, as $aϵ[6,100]$
$\Rightarrow a=6,12,20,30,42,56,72,90$
Thus, ${}^{\prime }{a}^{\prime }$ can attain $8$ different values.