## If a , b are roots of the equation x²-10cx-11d=0 and those of x²-10ax-11b=0 are c,d then find the value of a+b+c+d when a not equal to b no

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## Answers ( )

EXPLANATION.a, b are the roots of the equation =

=> x² – 10cx – 11d = 0 and x² – 10ax – 11b = 0 are

c and d

## To find the value of a + b + c + d

## when a ≠ b ≠ c ≠ d

## According to the question,

## => Conditions = 1

a, b are the roots of equation

=> x² – 10ax – 11b = 0

sum of zeroes of quadratic equation

=> a + b = -b/a

=> a + b = 10a ….. (1)

products of zeroes of quadratic equation

=> ab = c/a

=> ab = -11b …… (2)

From equation (2) we get,

=> a = -11

put the value of a = -11 in equation (1)

we get,

=> a + b = 10a

=> -11 + b = 10 ( -11 )

=> -11 + b = -110

=> b = -110 + 11

=> b = -99

## => conditions = 2

c, d are the roots of equation

=> x² – 10cx – 11d = 0

sum of zeroes of quadratic equation

=> c + d = -b/a

=> c + d = 10c ……. (3)

products of zeroes of quadratic equation

=> cd = c/a

=> cd = -11d …… (4)

From equation (4) we get,

=> c = -11

put the value of c = -11 in equation (3)

we get,

=> c + d = 10c

=> -11 + d = 10(-11)

=> -11 + d = -110

=> d = -99

## Therefore,

## A = -11 and B = -99

## C = -11 and D = -99

conditions are given

=> a ≠ b ≠ c ≠ d

=> -11 ≠ – 99 ≠ -11 ≠ -99

## Therefore,

value of = a + b + c + d

=> -11 + ( -99) + (- 11 ) + ( -99)

=> – 220

## values are = -220

We have,

…(1)

and

…(2)

Subtracting (2) from (1),

…(3)

Since a is the root of , we have,

…(4)

Since c is also a root of , hence,

…(5)

Subtracting (4) from (5), we get,

…(6)

[Assuming, . If , then from (1) and (2), we can conclude that , i.e., in this case, the given equation will be identical. So, I avoid this case.]

On adi\ding (1) & (2), we have,

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