Question

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 ??

Answer 1

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 2

Answer:

Hope it helps

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)

(Subtract) 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.

(a+b+c+d) = 1210