Question

Find the zeroes of p(x)=9x³-x and hence verify the relationship between zeroes and coefficients.​

Answer 1

Answer:

$\sf{The \ zeroes \ of \ the \ polynomial \ are}$

$\sf{0, \ \dfrac{1}{3} \ and \ -\dfrac{1}{3}.}$

Solution:

$\sf{p(x)=9x^{3}-x}$

$\sf{\therefore{9x^{3}-x=0}}$

$\sf{\therefore{x(9x^{2}-1)=0}}$

$\sf{\therefore{x=0 \ or \ 9x^{2}-1=0}}$

$\sf{\therefore{x=0 \ or \ x^{2}=\dfrac{1}{9}}}$

$\sf{\therefore{x=0 \ or \ x=\dfrac{1}{3} \ or \ x=-\dfrac{1}{3}}}$

$\sf\purple{\tt{\therefore{The \ zeroes \ of \ the \ polynomial \ are}}}$

$\sf\purple{\tt{0, \ \dfrac{1}{3} \ and \ -\dfrac{1}{3}.}}$

_______________________________

$\sf\blue{Relation \ between \ coefficients \ and \ zeroes}$

$\sf{The \ given \ cubic \ polynomial \ is}$

$\sf{p(x)=9x^{3}-x}$

$\sf{In \ standard \ form}$

$\sf{\leadsto{p(x)=9x^{3}+0x^{2}-x+0}}$

$\sf{Here, \ a=9, \ b=0, \ c=-1 \ and \ d=0}$

$\sf{Let \ \alpha=0,}$

$\sf{\beta=\dfrac{1}{3},}$

$\sf{\gamma=-\dfrac{1}{3}}$

______________________

$\sf{-\dfrac{b}{a}=-\dfrac{0}{9}}$

$\sf{\therefore{-\dfrac{b}{a}=0...(1)}}$

$\sf{\alpha+\beta+\gamma=\dfrac{1}{3}-\dfrac{1}{3}+0}$

$\sf{\therefore{\alpha+\beta+\gamma=0...(2)}}$

$\sf{from \ (1) \ and \ (2)}$

$\boxed{\sf{\alpha+\beta+\gamma=-\dfrac{b}{a}}}$

_______________________

$\sf{\dfrac{c}{a}=-\dfrac{1}{9}...(3)}$

$\sf{\alpha\beta+\beta\gamma+\alpha\gamma=(0)(\dfrac{1}{3})+(\dfrac{1}{3})(-\dfrac{1}{3})+(0)(-\dfrac{1}{3})}$

$\sf{\therefore{\alpha\beta+\beta\gamma+\alpha\gamma=-\dfrac{1}{9}...(4)}}$

$\sf{from \ (3) \ and \ (4)}$

$\boxed{\sf{\alpha\beta+\beta\gamma+\alpha\gamma=\dfrac{c}{a}}}$

_______________________

$\sf{-\dfrac{d}{a}=-\dfrac{0}{9}}$

$\sf{\therefore{-\dfrac{d}{a}=0...(5)}}$

$\sf{\alpha\beta\gamma=(0)(\dfrac{1}{3})(-\dfrac{1}{3})}$

$\sf{\therefore{\alpha\beta\gamma=0...(6)}}$

$\sf{from \ (5) \ and \ (6)}$

$\boxed{\sf{\alpha\beta\gamma=-\dfrac{d}{a}}}$

Answer 2

Answer:

The zeroes of the given polynomial are 0 1/3 and -1/3

Hope it helps you.