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

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

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Melanie 4 weeks 2021-08-17T17:00:17+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-17T17:01:49+00:00

    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

    0
    2021-08-17T17:02:11+00:00

    \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}}

    _____________________________

    hope it’s helpful

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