Quadratic EquationHard
Question
If the roots of the equation x2 - 2ax + a2 + a - 3 = 0 are real and less than 3, then
Options
A.
- 4 < a < 2
- 4 < a < 2
B.2 ≤ a ≤ 3
C.3 ≤ a ≤ 4
D.a > 4
Solution
Let f (x) = x2 - 2ax = a2 + a - 3
Since, both root are less than 3.
⇒ α < 3, β < 3
⇒ Sum, S = α + β < 6
⇒
< 3⇒
< 3⇒ α < 3 .......(i)
Again, product, P = αβ
⇒ P < 9
⇒ αβ < 9
⇒ α2 + a - 3 < 9
⇒ α2 + a - 12 < 0
⇒ (α - 3)(a + 4) < 0
⇒ -4 < a < 3 .......(ii) Agani, D = B2 - 4AC ≥ 0
⇒ (-2a)2 - 4.1 (a2 + a - 3) ≥ 0
⇒ 4a2 - 4a2 - 4a + 12 ≥ 0
⇒ - 4a + 12 ≥ 0
⇒ a ≤ 3 .......(iii)
Agani, a f (3) > 0
⇒ 1[(3)2 - 2a(3) + a2 + a - 3] > 0
⇒ 9 - 6a + a2 + a - 3 > 0
⇒ a2 - 5a + 6 > 0
⇒ (a - 2)(a - 3) > 0
∴ a ∈ ( - ∞, 2)
(3, ∞) .......(iv) From Eqs. (i),(ii),(iii) and (iv), we get
a ∈ (-4, 2)
Thereform, (a) is the answer.
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