Question

Find the zeroes of the quadratic polynomial a
6x^2-3-7x verify the
relationship between the zeroes and its coefficients,​

Answer 1

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

$\sf{\frac{3}{2} \ and \ \frac{-1}{3} \ are \ the \ zeroes}$

$\sf{of \ the \ polynomial.}$

$\sf\orange{Given:}$

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

$\sf{\implies{6x^{2}-3-7x}}$

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

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

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

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

$\sf{\implies{6x^{2}-3-7x}}$

$\sf{\implies{6x^{2}-7x-3}}$

$\sf{\implies{6x^{2}-9x+2x-3}}$

$\sf{\implies{3x(2x-3)+1(2x-3)}}$

$\sf{\implies{(2x-3)(3x+1)}}$

$\sf{\implies{x=\frac{3}{2} \ or \ \frac{-1}{3}}}$

$\sf\purple{\tt{\therefore{\frac{3}{2} \ and \ \frac{-1}{3} \ are \ the \ zeroes}}}$

$\sf\purple{\tt{of \ the \ polynomial.}}$

___________________________________

$\sf\blue{\underline{\underline{Verification:}}}$

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

$\sf{\implies{6x^{2}-3-7x}}$

$\sf{\implies{6x^{2}-7x-3}}$

$\sf{Here, \ a=6, \ b=-7 \ and \ c=3}$

$\sf{Let \ \frac{3}{2} \ be \ \alpha \ and \ \frac{-1}{3} \ be \ \beta}$

________________________________

$\sf{\alpha+\beta=\frac{3}{2}+(\frac{-1}{3})}$

$\sf{\therefore{\alpha+\beta=\frac{9-2}{6}}}$

$\sf{\therefore{\alpha+\beta=\frac{7}{6}...(1)}}$

$\sf{\frac{-b}{a}=\frac{-(-7)}{6}}$

$\sf{\therefore{\frac{-b}{a}=\frac{7}{6}...(2)}}$

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

$\sf{\alpha+\beta=\frac{-b}{a}}$

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

$\sf{\alpha\beta=\frac{3}{2}\times(\frac{-1}{3})}$

$\sf{\therefore{\alpha\beta=\frac{-3}{6}...(3)}}$

$\sf{\frac{c}{a}=\frac{-3}{a}...(4)}$

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

$\sf{\alpha\beta=\frac{c}{a}}$

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

$\sf{Hence, \ verified.}$

Answer 2

$\purple{\underline{\underline{\pink{\boxed{\boxed{\red{\star{\sf Question:}}}}}}}}$

$\rm{Find \ the \ zeroes \ of \ the \ quadratic \ polynomial}$

$\rm{6x^2-3-7x \ verify \ the \ relationship \ between}$

$\rm{the \ zeroes \ and \ its \ coefficients}$

_________________________________________

$\purple{\underline{\underline{\orange{\boxed{\boxed{\green{\star{\sf Answer:}}}}}}}}$

$\rm{\blue{\underline{\red{\frac{3}{2} \ and \ \frac{-1}{3} \ are \ the \ zeroes}}}}$

$\rm{\blue{\underline{\red{of \ the \ polynomial.}}}}$

_________________________________________

$\sf\large\underline\pink{Given:}$

$\rm{The \ given \ quadratic \ polynomial \ is}$

$\rm{\longrightarrow{6x^{2}-3-7x}}$

_________________________________________

$\sf\large\underline\blue{To \ Find}$

$\rm{The \ zeroes \ of \ the \ polynomial.}$

_________________________________________

$\purple{\underline{\underline{\blue{\boxed{\boxed{\red{\star{\sf Answer:}}}}}}}}$

$\rm{The \ given \ quadratic \ polynomial \ is}$

$\rm{\longrightarrow{6x^{2}-3-7x}}$

$\rm{\longrightarrow{6x^{2}-7x-3}}$

$\rm{\longrightarrow{6x^{2}-9x+2x-3}}$

$\rm{\lingrightarrow{3x(2x-3)+1(2x-3)}}$

$\rm{\longrightarrow{(2x-3)(3x+1)}}$

$\rm{\longrightarrow{x=\frac{3}{2} \ or \ \frac{-1}{3}}}$

$\rm{\blue{\underline{\red{\frac{3}{2} \ and \ \frac{-1}{3} \ are \ the \ zeroes}}}}$

$\rm{\blue{\underline{\red{of \ the \ polynomial.}}}}$

__________________________________________

$\pink{\underline{\underline{\green{\boxed{\boxed{\purple{\star{\sf Verification \ of \ the \ Answer:}}}}}}}}$

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

$\rm{\longrightarrow{6x^{2}-3-7x}}$

$\rm{\longrightarrow{6x^{2}-7x-3}}$

$\rm{Here, \ a=6, \ b=-7 \ and \ c=3}$

$\rm{Let \ \frac{3}{2} \ be \ \alpha \ and \ \frac{-1}{3} \ be \ \beta}$

________________________________

$\rm{\alpha+\beta=\frac{3}{2}+(\frac{-1}{3})}$

$\rm{\implies{\alpha+\beta=\frac{9-2}{6}}}$

$\rm{\implies{\alpha+\beta=\frac{7}{6}......(a)}}$

$\rm{\frac{-b}{a}=\frac{-(-7)}{6}}$

$\rm{\implies{\frac{-b}{a}=\frac{7}{6}...(b)}}$

$\rm{...from \ (a) \ and \ (b)}$

$\rm{\alpha+\beta=\frac{-b}{a}}$

$\rm{\implies{Sum \ of \ zeroes=\frac{-b}{a}}}$

$\rm{\alpha\beta=\frac{3}{2}\times(\frac{-1}{3})}$

$\rm{\implies{\alpha\beta=\frac{-3}{6}.....(c)}}$

$\rm{\frac{c}{a}=\frac{-3}{a}......(d)}$

$\rm{...from \ (c) \ and \ (d)}$

$\rm{\alpha\beta=\frac{c}{a}}$

$\rm{\therefore{Product \ of \ zeroes=\frac{c}{a}}}$

$\rm\orange{Hence, \ Answer \ is \ verified}$

___________________________________________

$\rm\blue{Hope \ it \ helps \ :))}$