Question

If the sum of the zeroes of the polynomial p(x)=2x^2-6x+k is half of its product of its zeroes then find value of k​

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ value \ of \ k \ is \ 12.}$

$\sf\orange{Given:}$

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

$\sf{\implies{p(x)=2x^{2}-6x+k}}$

$\sf{\implies{The \ sum \ of \ the \ zeroes \ is \ half}}$

$\sf{of \ it's \ product. }$

$\sf\pink{To \ find:}$

$\sf{The \ value \ of \ k.}$

$\sf\green{\underline{\underline{Solution:}}}$

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

$\sf{\implies{p(x)=2x^{2}-6x+k}}$

$\sf{Here \ a=2, \ b=-6 \ and \ c=k}$

$\sf{Sum \ of \ zeroes=\frac{-b}{a}=\frac{6}{3}}$

$\sf{\therefore{Sum \ of \ zeroes=3}}$

$\sf{Product \ of \ zeroes=\frac{c}{a}=\frac{k}{2}}$

$\sf{According \ to \ the \ given \ condition}$

$\sf{\frac{1}{2}\times\frac{k}{2}=3}$

$\sf{\therefore{k=3\times4}}$

$\sf{\therefore{k=12}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ k \ is \ 12.}}}$