Quadratic EquationHard
Question
If roots of x2 - (a - 3) x + a = 0, are such that at least one of them is greater than 2, then :-
Options
A.a ∈ [7, 9]
B.a ∈ [7, ∞)
C.a ∈ [9, ∞)
D.a ∈ [7, 9)
Solution
x2 - (a - 3) x + a = 0
two possibler : -
Case-1 one root > 2
f(2) < 0
4 - (a - 3) × 2 + a < 0
a > 10
Case-2 D ≥ 0 (for real roots)
(a - b)2 - 4a ≥ 0
a2 - 10a + a ≥ 0
a ∈ (-∞, 1] ∪ [a, ∞)
⇒ bot roots greaser then 2
f(2) ≥ 0 → a ≤ 10
⇒ minima i.e. -
> 2
+
> 2
a > 1
one solution will be union of both cases and D ≥ 0
so a ∈ [a, 10) ∪ (10, ∞)
a ∈ [a, ∞)
two possibler : -
Case-1 one root > 2
f(2) < 0
4 - (a - 3) × 2 + a < 0
a > 10
Case-2 D ≥ 0 (for real roots)
(a - b)2 - 4a ≥ 0
a2 - 10a + a ≥ 0
a ∈ (-∞, 1] ∪ [a, ∞)
⇒ bot roots greaser then 2
f(2) ≥ 0 → a ≤ 10
⇒ minima i.e. -
> 2+
> 2a > 1
one solution will be union of both cases and D ≥ 0
so a ∈ [a, 10) ∪ (10, ∞)
a ∈ [a, ∞)
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