Total number of values of -a- so that x 2- x - a -0 has integral roots- a - N - 5- a -99 isA- 2B- 4C- 5D- 8

The correct option is D 8 x^2 x a = 0 ∴ Δ =1 4( a) ⇒Δ =1 + 4a = An odd number i.e Δ =(1+2k)^2 (Because if root is an integer, Discriminant must be a ...

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Standard XII

Mathematics

Discriminant

Total number ...

Question

Total number of values of 'a' so that $x^{2}$ - x - a = 0 has integral roots, $a ϵ N$ & 5 < a < 99 is

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Solution

The correct option is D
8

$x^{2}$ - x - a = 0
$∴ Δ = 1 - 4 (- a) \Rightarrow Δ = 1 + 4 a$ = An odd number
$i . e Δ = (1 + 2 k)^{2} (Because if root is an integer, Discriminant must be a perfect square) i . e 1 + 4 a = 1 + 4 k^{2} + 4 k \Rightarrow a = k (k + 1)$

$∴$ 8 cases only

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Total number of values of 'a' so that x2 - x - a = 0 has integral roots, aϵN & 5 < a < 99 is

The correct option is D 8 x2 - x - a = 0 ∴ Δ=1−4(−a) ⇒Δ=1+4a = An odd number i.e Δ=(1+2k)2 (Because if root is an integer, Discriminant must be a perfect square) i.e 1+4a=1+4k2+4k⇒ a=k(k+1) ∴ 8 cases only

Total number of values of 'a' so that $x^{2}$ - x - a = 0 has integral roots, $a ϵ N$ & 5 < a < 99 is

The correct option is D
8

$x^{2}$ - x - a = 0
$∴ Δ = 1 - 4 (- a) \Rightarrow Δ = 1 + 4 a$ = An odd number
$i . e Δ = (1 + 2 k)^{2} (Because if root is an integer, Discriminant must be a perfect square) i . e 1 + 4 a = 1 + 4 k^{2} + 4 k \Rightarrow a = k (k + 1)$

$∴$ 8 cases only