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Let f(x)=2x^2+3px-2q and g(x)=x^2+q,where p and q are constants.It is given that x-a is a common factor of f(x) and g(x) ,where p ,q and a are non-zero constants.Show that 9p^2+16q=0 .

  • Math -

    Let f(x) = 2x² + 3px - 2q = 0 and g(x) = x² + q = 0
    Both have common factor (x - a)
    Hence by factor theorem, f(a) = g(a) = 0

    ==> g(a): a² + q = 0; so q = -a² ------ (1)

    and f(a): 2a² + 3pa - 2q = 0
    Substituting q = -a² from (1) in the above,
    4a² + 3pa = 0
    Factorizing, a(4a + 3p) = 0
    ==> Either a = 0 or a = -3p/4

    But given a is a non zero constant; hence a = 0 is rejected.
    Thus a = -3p/4 only

    Substituting this value in (1), q = -9p²/16
    ==> 16q = -9p²

    ==> 9p² + 16q = 0 [Proved]

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