Question

Let r be a root of the equation x^2 +2x+6=0. Then sum of digits in the value of
(r + 2) (r +3) (r + 4) (+5)
is equal to ​

Answer 1

Step-by-step explanation:

Given Let r be a root of the equation x^2 +2x+6=0. Then sum of digits in the value of  (r + 2) (r +3) (r + 4) (+5)

is equal to ​

• Now the given equation is x^2 + 2x + 6 = 0
• Then we can write as r^2 + 2r + 6 = 0
• Now we have r^2 = - 2r – 6 and  
•               So r^2 + 2r = - 6

• We need to find (r + 2) (r + 3) (r + 4) (r + 5)
•                     = (r^2 + 5r + 6) (r^2 + 9r + 20)  
•                   = (- 2r – 6 + 5r + 6) (- 2r – 6 + 9r + 20) (since r^2 = - 2r – 6)
•                   = 3r (7r + 14)
•                   = 3r x 7 (r + 2)
•                     = 21 r (r + 2)
•                    = 21 (r^2 + 2r)
•                    = 21 (- 6)  (since r^2 + 2r = - 6)
•                   = - 126

Reference link will be

https://brainly.in/question/7202357