Question

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 not equal to c not equal to d

Answer 1

EXPLANATION.

• GIVEN

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

Answer 2

$\Large\underline\mathbb\pink{ANSWER$

We have,

$a+b=10c$...(1)

and

$c+d=10a$...(2)

Subtracting (2) from (1),

$(a-c)+(b-d)=10(c-a)$

$\therefore b-d=11(c-a)$...(3)

Since a is the root of $x^{2}-10cx-11d=0$, we have,

$a^{2} -10ac-11d=0$...(4)

Since c is also a root of $x^{2} -10ax-11b=0$, hence,

$c^{2} -10ca-11b=0$...(5)

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

$c^{2} -a^{2} =11(b-d)$

$\implies (c+a)(c-a)=11\times11(c-a)$

$\therefore c+a=121$...(6)

[Assuming, $c\neq a$. If $c=a$, then from (1) and (2), we can conclude that $b=d$, i.e., in this case, the given equation will be identical. So, I avoid this case.]

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

$a+b+c+d=10(c+a)=10\times121 [from 6]$

$\boxed{\red{a+b+c+d=1210}}$

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