Question

Find the zeroes of the quadractic polynomial 6x²-7x-3 and verify the relationship between the zeroes & the coefficients.

Answer 1

AnswEr :

$\star \; \small\bold{\underline{\underline{\sf{\pink{Given \: polynomial : 6x^2 - 7x - 3}}}}}$

$:\implies\sf 6x^2 - 7x - 3 = 0 \\\\\\:\implies\sf 6x^2 - 9x +2x - 3 = 0 \\\\\\:\implies\sf 3x \Big(2x - 3 \Big) + 1 \Big(2x - 3 \Big) = 0\\\\\\:\implies\sf \Big(3x + 1 \Big) \Big(2x - 3 \Big) = 0\\\\\\:\implies\sf 3x + 1 = 0\\\\\\:\implies\sf 3x = - 1\\\\\\:\implies\sf x = \dfrac{-1}{\: 3} \;\:\;\;\; \& \: \: \: \: 2x - 3 = 0\\\\\\:\implies\sf 2x = 3 \\\\\\:\implies\sf x = \dfrac{3}{2}$

$:\implies\sf \alpha = \dfrac{-1}{\:3} \:\:\&\;\; \beta = \dfrac{3}{2}$

$\rule{150}2$

$\star \:\:\small\bold{\underline{\sf{\pink{ Relationship \: b/w \: Zeroes \: \& \: Coefficients}}}}$

$\underline{\dag\:\boldsymbol{Sum \: of \: Zeroes \: : }}$

$\dashrightarrow{\underline{\boxed{\sf{\purple{Sum \: of \: Zeroes = \alpha \:+\: \beta = \dfrac{-b}{a}}}}}}$

$:\implies\sf \bigg(\dfrac{-1}{\:3} + \dfrac{3}{2} \bigg) \\\\\\:\implies\sf \bigg(\dfrac{7}{6} = \dfrac{-b}{a} \bigg)$

$\rule{150}2$

$\underline{\dag\:\boldsymbol{Product \: of \:Zeroes \: }}$

$\dashrightarrow{\underline{\boxed{\sf{\purple{Product \: of \: Zeroes = \alpha \:\times \: \beta = \dfrac{c}{a}}}}}}$

$:\implies\sf \bigg( \dfrac{-1}{3} \times \dfrac{3}{2} \bigg) \\\\\\:\implies\sf \bigg(\dfrac{-3}{6} = \dfrac{ -1}{\: 2} = \dfrac{c}{a} \bigg)$

Answer 2

Answer:

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