\xa0if one zero of polynomial 2x^2-(3k-1) x-9 is the\xa0negative of the other find K and the zeros.

\xa0if one zero of polynomial 2x^2-(3k-1) x-9 is the\xa0negative of th

1.	\xa0if one zero of polynomial 2x^2-(3k-1) x-9 is the\xa0negative of the other find K and the zeros.
Answer» Also find the zeros Ans. Polynomial : 2x2 - (3k-1) x - 9\xa0on comparing with ax2\xa0+ bx + c, we get, a = 2, b = -(3k-1) and c = -9Let one zero =\xa0{tex}\\alpha {/tex},\xa0then second zero ll be =\xa0{tex}- \\alpha {/tex}We knowSum\xa0of zeroes =\xa0{tex}-{b\\over a}{/tex}=>\xa0{tex}-\\alpha + \\alpha = {-[{-({3k-1)}}]\\over 2}{/tex}=>\xa0{tex}0 = {3k-1\\over 2}{/tex}=> 3k -1 = 0\xa0=> 3k = 1\xa0=> k =\xa0{tex}{1\\over 3}{/tex}