18.
If a be the ratio of the roots of the quadratic equation in x, 3m2x2 + m(m – 4)X + 2 = 0, then the least
value of m for which a
= 1, is :
[JEE(Main) 2019, Online (12-01-19),P-1 (4, -1), 120]
(1)-2 + 2
(2) 4-312
(3) 2-13
(4)4-2,73
Answers
Answer:
m = 4 ± 2√6 ( as per your question) a = 1
4 - 3√2 ( as per JEE exam) when a + 1/a = 1
Step-by-step explanation:
3m²x² + m(m-4)x + 2 = 0
a is the ratio of roots & a = 1
=> Roots are Equal
Roots would be equal if D = 0
=> (m(m-4))² - 4*3m²*2 = 0
Dividing by m²
=> (m-4)² - 24 = 0
=> m² + 16 - 8m - 24 = 0
=> m² - 8m - 8 = 0
=> m = (8 ± √8² + 32)/2
=> m = 4 ± 2√6
3m²x² + m(m-4)x + 2 = 0
Actual Question as per JEE exam is a + 1/a = 1
let say α & β are roots
α/β + β/α = 1
=> (α² + β²) = αβ
=> (α + β)² - 2αβ = αβ
=> (α + β)² = 3αβ
α + β = -m(m-4)/3m² αβ = 2/3m²
=> (-m(m-4)/3m²)² = 3* 2/3m²
=> m²(m-4)²/3m² = 6
=> (m-4)² = 18
=> m² + 16 - 8m = 18
=> m² - 8m - 2 = 0
=> m = (8 ± √(64 + 8))/2 = 4 ± 3√2
4 - 3√2 is one of the given option
Step-by-step explanation:
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