Question

Exactly one root of the equation x2 - (a + 3)x + 4 = 0 lies between 3 and 4 for a E(m,n)
then (m + n) is​

Answer 1

Given : Exactly one root of the equation x²  - (a + 3)x  + 4 = 0 lies between 3 and 4 for a ∈ (m , n)

To Find : m + n

Solution:

x²  - (a + 3)x  + 4 = 0

=> x = ( (a + 3)  ± √(a + 3)² - 4(4) )/2

=> x =  ( (a + 3)  ± √a² + 6a - 7)/2

a² + 6a - 7 = (a + 7)(a - 1)

(a + 7)(a - 1) ≥ 0

=> a≥ 1  , a ≤ - 7

x²  - (a + 3)x  + 4 = 0

one root lies between 3 and 4

Product of roots is 4

Hence another root  lies  between  1 & 4/3

Sum of roots  =  a + 3

Range of Sum of roots = 4 + 1 = 5   or  3 + 4/3  =  13/3

13/3  <  a + 3   <  5

=> 4/3 <  a   <  2

a ∈ ( 4/3 , 2 )

m = 4/3

n = 2

4/3 + 2  = 10/3

m + n   = 10/3

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