Math, asked by angelachawngthu21, 7 months ago

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

Answers

Answered by Anonymous
6

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}}}

Answered by Anonymous
2

Answer:

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

Hope it helps you.

Similar questions