please solve this Question : If x2 – 10ax -11b = 0 has ‘c’ and ‘d’ as its roots and the equation x2 – 10cx -11d = 0 has its roots a and b , then find the value of a+b+c+d ??
Answers
Answer :---
we are Given
: x2 – 10ax -11b = 0
And x2 – 10ax -11b = 0
By using the concepts of sum of roots and products.
a+b = 10c
&
c+d = 10a
So,
it becomes (a-c) + (b-d) = 10(c-a)
This gives,
⭐(b-d) = 11(c-a) ….. (1)
Since, ‘c’ is a root of x2 – 10ax -11b = 0
Hence,
⭐c2 – 10ac -11b = 0 …..(2)
Similarly,
‘a’ is a root of the equation x2 – 10cx -11d = 0
So,
⭐a2 – 10ca -11d = 0 _____(3)
Now, ➖ (minise) equation (3) from (2), Value
(c2 - a2) = 11(b-d)_____(4)
•°•, (c+a) (c-a) = 11.11(c-a) ____ (From eq(1))
Hence, c+a = 121
Therefore, a+b+c+d = 10c + 10a
= 10(c+a)
= 1210.
•°• the required value of (a+b+c+d) = 1210
Answer :---
we are Given
: x2 – 10ax -11b = 0
And x2 – 10ax -11b = 0
By using the concepts of sum of roots and products.
a+b = 10c
&
c+d = 10a
So,
it becomes (a-c) + (b-d) = 10(c-a)
This gives,
⭐(b-d) = 11(c-a) ….. (1)
Since, ‘c’ is a root of x2 – 10ax -11b = 0
Hence,
⭐c2 – 10ac -11b = 0 …..(2)
Similarly,
‘a’ is a root of the equation x2 – 10cx -11d = 0
So,
⭐a2 – 10ca -11d = 0 _____(3)
Now, ➖ (minise) equation (3) from (2), Value
(c2 - a2) = 11(b-d)_____(4)
•°•, (c+a) (c-a) = 11.11(c-a) ____ (From eq(1))
Hence, c+a = 121
Therefore, a+b+c+d = 10c + 10a
= 10(c+a)
= 1210.
•°• the required value of (a+b+c+d) = 1210