Math, asked by SaivardhanJ, 7 months ago

(x^(2)+x+2)^(2)-(a-3)(x^(2)+x+1)(x^(2)+x+2)+(a-4)(x^(2)+x+1) has at least one real root then 'a' could lie anywhere in the interval(s) ​

Answers

Answered by amitnrw
0

Given :  (x² + x + 2)²  - (a - 3)(x² + x + 1)(x² + x + 2) + (a - 4)(x² + x + 1)²  has atleast one real root

To Find : 'a' could lie anywhere in the interval(s) ​

Solution:

(x² + x + 2)²  - (a - 3)(x² + x + 1)(x² + x + 2) + (a - 4)(x² + x + 1)²  = 0

=> (x² + x + 2)² - (a - 4)(x² + x + 1)(x² + x + 2)  - 1(x² + x + 1)(x² + x + 2) + (a - 4)(x² + x + 1)²  = 0

=>    (x² + x + 2)((x² + x + 2) - (a - 4)(x² + x + 1)) - (x² + x + 1)((x² + x + 2) -  - (a - 4)(x² + x + 1) =0

=> (x² + x + 2) - (a - 4)(x² + x + 1)(  (x² + x + 2) -  (x² + x + 1))  = 0

=> (x² + x + 2) - (a - 4)(x² + x + 1) 3  = 0

=> (x²(1 - a+ 4)  + x( 1 - a + 4) + 2 - a + 4  = 0

=> x²(5 - a)  + x(5 - a) + 6-a  = 0  

will have real root  if  D ≥ 0 => b² - 4ac ≥ 0

=>  ( 5 - a)²  - 4(5-a)(6 - a) ≥ 0

=> (5 - a) (5 - a  - 24 + 4a ) ≥ 0

=> (5 - a) ( 3a - 19) ≥ 0

=> (a - 5)(3a - 19) ≤ 0

=>    5 ≤ a ≤ 19/3

'a' could lie anywhere in the interval(s) ​   5 ≤ a ≤ 19/3

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