Total number of values of 'a' , so that x2-x-a=0 has integral roots, where a∈ N and 6 le a le 100 , is equal to

Question

Total number of values of 'a' , so that x2-x-a=0 has integral roots, where a∈ N and 6 le a le 100 , is equal to (A)  2  (B)  4  (C)  8  (D)  6

Tardigrade Admin · Accepted Answer

x2-x-a=0, D=1+4 a= odd D must be perfect square of some odd integer. Let D=(2 λ+1)2 ⇒ 1+4 a=1+4 λ2+4 λ ⇒ a=λ(λ+1), as a ∈[6,100] ⇒ a=6,12,20,30,42,56,72,90 Thus, ' a prime can attain 8 different values.

Q. Total number of values of $^{'} a^{'}$ , so that $x^{2} - x - a = 0$ has integral roots, where $a \in N$ and $6 \leq a \leq 100$ , is equal to

$2$

$4$

$8$

$6$

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

Q. Total number of values of ′a′ , so that x2−x−a=0 has integral roots, where a∈N and 6≤a≤100 , is equal to

2

4

8

6

x2−x−a=0,D=1+4a= odd D must be perfect square of some odd integer. Let D=(2λ+1)2⇒1+4a=1+4λ2+4λ⇒a=λ(λ+1), as a∈[6,100]⇒a=6,12,20,30,42,56,72,90 Thus, ' a′ can attain 8 different values.

Q. Total number of values of $^{'} a^{'}$ , so that $x^{2} - x - a = 0$ has integral roots, where $a \in N$ and $6 \leq a \leq 100$ , is equal to

$2$

$4$

$8$

$6$

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