Question

find zero of all the following quadratic polynomial and verify the relationship between zero and the coefficient 6√3x2-7x-√3​

Answer 1

Answer:

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

$\sf{-\dfrac{1}{3\sqrt3} \ and \ \dfrac{\sqrt3}{2}.}$

Given:

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

$\sf{p(x)=6\sqrt3x^{2}-7x-\sqrt3.}$

To find:

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

Solution:

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

$\sf{p(x)=6\sqrt3x^{2}-7x-\sqrt3}$

$\sf{p(x)=0}$

$\sf{\therefore{6\sqrt3x^{2}-7x-\sqrt3=0}}$

$\sf{\therefore{6\sqrt3x^{2}-9x+2x-\sqrt3=0}}$

$\sf{\therefore{3\sqrt3x(2x-\sqrt3)+1(2x-\sqrt3)=0}}$

$\sf{\therefore{(3\sqrt3x+1)(2x-\sqrt3)=0}}$

$\sf{\therefore{x=-\dfrac{1}{3\sqrt3} \ or \ \dfrac{\sqrt3}{2}}}$

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

$\sf\purple{\tt{-\dfrac{1}{3\sqrt3} \ and \ \dfrac{\sqrt3}{2}.}}$

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Verification:

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

$\sf{p(x)=6\sqrt3x^{2}-7x-\sqrt3.}$

$\sf{Here, \ a=6\sqrt3, \ b=-7 \ and \ c=-\sqrt3}$

$\sf{Sum \ of \ zeroes=-\dfrac{1}{3\sqrt3}+\dfrac{\sqrt3}{2}}$

$\sf{\therefore{Sum \ of \ zeroes=\dfrac{-2+9}{6\sqrt3}}}$

$\sf{\therefore{Sum \ of \ zeroes=\dfrac{7}{6\sqrt3}...(1)}}$

$\sf{-\dfrac{b}{a}=\dfrac{7}{6\sqrt3}...(2)}$

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

$\boxed{\sf{Sum \ of \ zeroes=-\dfrac{b}{a}}}$

$\sf{Product \ of \ zeroes=-\dfrac{1}{3\sqrt3}\times\dfrac{\sqrt3}{2}}$

$\sf{\therefore{Product \ of \ zeroes=-\dfrac{\sqrt3}{6\sqrt3}...(3)}}$

$\sf{\dfrac{c}{a}=-\dfrac{\sqrt3}{6\sqrt3}...(4)}$

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

$\boxed{\sf{Product \ of \ zeroes=\dfrac{c}{a}}}$

$\sf{Hence, \ verified.}$