If one root the equation ax²+ bx + c = 0 is three times times the other, then b²: ac =
(a)3 : 1
(b)3 : 16
(c)16 : 3
(d)16 : 1
Answers
SOLUTION :
Option (c) is correct : 16 : 3
Let α be the one root of the equation and β be the other root .
Given : ax² + bx + c = 0 , α = 3β
On comparing the given equation with ax² + bx + c = 0
Here, a = a , b = b , c = c
Sum of zeroes = - b/a
α + β = - b/a
3β + β = - b/a
4β = - b/a
β = - b/4a ………………(1)
Product of zeroes = c/a
α × β = c/a
3β × β = c/a
3β² = c/a
β² = c/3a ……………..(2)
Put the value of β = - b/4a in eq 2,
(- b/4a)² = c/3a
b²/16a² = c/3a
b² = (c/3a) × 16a²
b² = 16ac/3
b²/ac = 16/3
b² : ac = 16 : 3
Hence, b² : ac is 16 : 3
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Let α be the one root of the equation and β be the other root .
Given : ax² + bx + c = 0 , α = 3β
On comparing the given equation with ax² + bx + c = 0
Here, a = a , b = b , c = c
Sum of zeroes = - b/a
α + β = - b/a
3β + β = - b/a
4β = - b/a
β = - b/4a ………………(1)
Product of zeroes = c/a
α × β = c/a
3β × β = c/a
3β² = c/a
β² = c/3a ……………..(2)
Put the value of β = - b/4a in eq 2,
(- b/4a)² = c/3a
b²/16a² = c/3a
b² = (c/3a) × 16a²
b² = 16ac/3
b²/ac = 16/3
b² : ac = 16 : 3
Hence, b² : ac is 16 : 3