Question

find the zeroes of the quadratic polynomial 6x2-3-7x and verify relationship between the zeroes and their coefficients ​

Answer 1

Answer:

$6x^2 - 7x - 3 = 0\\\\6x^2 - 9x + 2x - 3 =0\\\\3x(2x-3) + 1(2x - 3) = 0\\\\(3x + 1)(2x - 3) =0\\\\3x + 1 =0 \ \ \ \ \ OR \ \ \ \ \ 2x - 3 = 0\\\\3x = 1 \ \ \ \ \ \ \ \ \ \ OR \ \ \ \ \ 2x = 3\\\\x = \frac{1}{3} \ \ \ \ \ \ \ \ \ \ \ OR \ \ \ \ \ \ x = \frac{3}{2}\\\\Thus \ the \ zeroes \ are \ \frac{1}{3} \ and \ \frac{3}{2}$

Step-by-step explanation:

Answer 2

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

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

$\sf{\therefore\; Zeroes\; of \; the \: Given \; polynomial \; are \; \dfrac{3}{2} \:\&\; \dfrac{-1 \: }{\; \: 3 \: }.}$

$\rule{250px}{.3ex}$

$\bf{\dag} \: \: \underline{\textsf{Relation b/w Coefficients \& Zeroes \: :}}$⠀⠀

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${\qquad\maltese\:\:\textsf{Sum of Zeroes :}} \\\\\dashrightarrow\sf\:\:\alpha +\beta= \dfrac{ - \:b \: \: \: }{ \: \: \: a \: \: \:}\\\\\\\dashrightarrow\sf \bigg(\dfrac{3}{2}\bigg) + \bigg(\dfrac{-1}{ \: \: 3}\bigg) = - \dfrac{-7}{ \: \: 6} \\\\\\\dashrightarrow{\underline{\boxed{\frak{\dfrac{7}{6} = \dfrac{7}{6}}}}}$

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${\qquad\maltese\:\:\textsf{Product of Zeroes :}}\\\\\dashrightarrow\sf\:\:\alpha\beta=\dfrac{c}{a}\\\\\\\dashrightarrow\sf \bigg(\dfrac{3}{2}\bigg) \times \bigg(\dfrac{-1 \: \: }{ \: \: 3 \: \: }\bigg) = \dfrac{-3 \: \: }{ \: \: 6 \: \: } \\\\\\\dashrightarrow{\underline{\boxed{\frak{\dfrac{-1}{\;2} = \dfrac{-1}{\;2}}}}}$

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$\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Verified!}}}}$