posted by Anonymous .
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 .
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]