Question

Find the zeros of the quadratic polynomial 2 root 3 x square minus 5 x plus root 3 and verify the relationship between the zeros and the coefficients.

Answer 1

$\star\;\normalsize{\underline{\underline{\frak \pink{GivEn:-}}}}$

• $\normalsize\sf p(x) = 2 \sqrt{3}x^2 - 5x + \sqrt{3}$

$\star\;\normalsize{\underline{\underline{\frak \pink{To\;Find:-}}}}$

• Zeroes of polynomial and to verify relationship b/w zeroes and the coefficient.

$\star\;\normalsize{\underline{\underline{\frak \red{Solution:-}}}}$

$\;\;\;\;\;\;\;\;\;\normalsize\sf {\underline{2 \sqrt{3}x^2 - 5x + \sqrt{3} = 0}} \\\\ :\implies \normalsize\sf 2 \sqrt{3}x^2 - 2x - 3x + \sqrt{3} = 0 \\\\ :\implies \normalsize\sf 2 \sqrt{3}x^2 - 2x - 3x + \sqrt{3} = 0 \\\\ :\implies \normalsize\sf 2x( \sqrt{3}x - 1) - \sqrt{3}( \sqrt{3}x - 1) \\\\ :\implies \normalsize\sf (2x - \sqrt{3})( \sqrt{3}x - 1) \\\\ :\implies \normalsize\sf either\; (2x - \sqrt{3}) or ( \sqrt{3}x - 1) \; is\;equal\;to\;zero \\\\ :\implies \normalsize\sf \therefore x = \dfrac{ \sqrt{3}}{2} and \dfrac{1}{ \sqrt{3}}$

$\normalsize\sf Let's \; \dfrac{ \sqrt{3}}{2}\;be\; \alpha \;and\; \dfrac{ \sqrt{1}}{ \sqrt{3}}\;be\; \beta.$

$\rule{200}{3}$

$\star\;\normalsize{\underline{\underline{\frak \purple{Verification:-}}}}$

• Relationship b/w zeroes and the coefficient.

$\star\;\normalsize\sf {\underline{Sum\;of\;zeroes\;( \alpha + \beta):-}} \; \bf{ \dfrac{-b}{a}} \\\\ \leadsto \normalsize\sf \bigg( \dfrac{ \sqrt{3}}{2} + \dfrac{1}{ \sqrt{3}} \bigg) = - \bigg( \dfrac{-5}{2 \sqrt{3}} \bigg) \\\\ \leadsto \normalsize\sf \dfrac{3 + 2}{2 \sqrt{3}} = \dfrac{5}{2 \sqrt{3}} \\\\ \leadsto \normalsize\sf \underbrace{ \dfrac{5}{2 \sqrt{3}} = \dfrac{5}{2 \sqrt{3}}}_{LHS = RHS}$

$\star\;\normalsize\sf {\underline{Product\;of\;zeroes\;( \alpha \beta):-}} \; \bf{ \dfrac{c}{a}} \\\\ \leadsto \normalsize\sf \bigg( \dfrac{ \sqrt{3}}{2} \times \dfrac{1}{ \sqrt{3}} \bigg) = \dfrac{1}{2} \\\\ \leadsto\normalsize\sf \dfrac{ \cancel{ \sqrt{3}}}{2 \cancel{ \sqrt{3}}} = \dfrac{1}{2} \\\\ \leadsto\normalsize\sf \underbrace{ \dfrac{1}{2} = \dfrac{1}{2}}_{LHS = RHS}$

$\normalsize{\underline{\underline{\sf{\red{\dag\; Hence\; Verified!!}}}}}$

$\rule{200}{3}$