Question

If a and b are the zeros of polynomial 6x²-(k+1)-7x such that one zero is 11/6 more than the other, find the value of k.

Answer 1

$\textbf{Given: a and b are zeros of the polynomial}\;6x^2-7x-(k+1)$

$\text{Then, }$

$\text{Sum of the zeros=}\;a+b=\frac{7}{6}$ ......(1)

$\text{Product of the zeros=}\;ab=\frac{-(k+1)}{6}$

$\text{Also, }\;a=\frac{11}{6}+b$ ......(2)

$\text{Using (2) in (1), we get}$

$\frac{11}{6}+b+b=\frac{7}{6}$

$\implies\;2b=\frac{7}{6}-\frac{11}{6}$

$\implies\;2b=-\frac{4}{6}$

$\implies\;b=\frac{-2}{6}$

$\implies\boxed{\bf\;b=\frac{-1}{3}}$

$\text{(2)}\implies\;a=\frac{11}{6}-\frac{1}{3}$

$\implies\;a=\frac{11-2}{6}$

$\implies\;a=\frac{9}{6}$

$\implies\boxed{\bf\;a=\frac{3}{2}}$

$\text{consider,}$

$ab=\frac{-(k+1)}{6}$

$\implies\,(\frac{3}{2})(\frac{-1}{3})=\frac{-(k+1)}{6}$

$\implies\,\frac{-1}{2}=\frac{-(k+1)}{6}$

$\implies\,1=\frac{k+1}{3}$

$\implies\,k+1=3$

$\implies\boxed{\bf\,k=2}$

Find more:

Obtain other zeroes of the polynomial

f(x) = 2x4 + 3x3 - 5x2 - 9x - 3

if two of its zeroes are √3 and - √3.​

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