Question

Find the zeroes of the given quadratic polynomial ² + ⅙ − 2 and verify the relationship

between the zeroes and the coefficient​

Answer 1

Given Polynomial : x² + ⅙x - 2

We've to find relationship b/w zeroes and Coefficient of given quadratic Polynomial.

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☆ Let's find out zeroes of Given Polynomial :

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$\qquad:\implies\sf x^2 + \dfrac{1}{6}\:x - 2 = 0\\\\\\ \qquad:\implies\sf 6x^2 + x - 12 = 0\\\\\\ \qquad:\implies\sf 6x^2 - 8x + 9x - 12 = 0\\\\\\ \qquad:\implies\sf 2x(3x - 4) + 3(3x - 4) = 0\\\\\\ \qquad:\implies\sf (3x - 4)(2x + 3) = 0\\\\\\ \qquad:\implies{\underline{\boxed{\pmb{\frak{\red{x = 4/3\:\:or\:\:x = - 3/2}}}}}}\:\bigstar$

$\therefore\:{\underline{\sf{Hence,\:The\:zeroes\:of\:Polynomial\:are\:{\pmb{4/3}}\:{\sf{\&}}\:{\bf{- 3/2}}.}}}$

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✇ Let's consider α and β be zeroes of Polynomial.

Here, In the given Polynomial x² + ⅙x - 2, {a = 1 , b = 1/6 & c = - 2}.

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☆ Now, Let's verify the relationship between zeroes and Coefficient :

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$\bf{\dag}\:{\underline{\boxed{\pmb{\sf{Sum\:of\:zeroes\:\purple{(\alpha + \beta)}\::}}}}}$

$\qquad\quad\dashrightarrow\sf \alpha + \beta = \dfrac{-b}{a}\\\\\\ \qquad\quad\dashrightarrow\sf \bigg(\dfrac{4}{3}\bigg) + \bigg(\dfrac{-3}{2}\bigg) = \dfrac{-\frac{1}{6}}{1}\\\\\\ \qquad\quad\dashrightarrow\sf \dfrac{4}{3} - \dfrac{3}{2} = \dfrac{-1}{6}\\\\\\ \qquad\quad\dashrightarrow{\boxed{\boxed{\frak{\pink{\dfrac{-1}{6} =\dfrac{-1}{6}}}}}}$

$\bf{\dag}\:{\underline{\boxed{\pmb{\sf{Product\:of\:zeroes\:\purple{(\alpha \beta)}\::}}}}}$

$\qquad\quad\dashrightarrow\sf \alpha \beta = \dfrac{c}{a}\\\\\\ \qquad\quad\dashrightarrow\sf \bigg(\dfrac{4}{3}\bigg) \bigg(\dfrac{-3}{2}\bigg) = \dfrac{-2}{1}\\\\\\ \qquad\quad\dashrightarrow\sf \cancel{\dfrac{- 12}{6}} = - 2\\\\\\ \qquad\quad\dashrightarrow{\boxed{\boxed{\frak{\pink{-2 = -2}}}}}$

Hence Verified !

Answer 2

Given :-

Quadratic polynomial x² + ⅙ − 2

To Find :-

Zeroes

Solution :-

Finding zeroes

$\sf x^{2} +\dfrac{1}{6}x - 2 = 0$

$\sf x^{2} + \dfrac{x}{6} - 2 =0$

$\sf 6 \times x^2 + x - 2 = 0$

$\sf 6x^2 + x - 2=0$

$\sf 6x^2 - 8x +9x- 12=0$

$\sf 2x(3x-4)+3(3x-4)=0$

$\sf (3x-4)(2x+3)=0$

Either

$\sf 3x - 4 = 0$

$\sf 3x = 4$

$\sf x = \dfrac{4}{3}$

or

$\sf 2x + 3 =0$

$\sf 2x=-3$

$\sf x=\dfrac{-3}{2}$

Verification

Sum of zeroes = -b/a

$\sf\dfrac{4}{3} + \dfrac{-3}{2} = \dfrac{-1}{6}$

$\sf\dfrac{8 +(-9)}{6} = \dfrac{-1}{6}$

$\sf\dfrac{8 -9}{6} = \dfrac{-1}{6}$

$\sf\dfrac{-1}{6} = \dfrac{-1}{6}$

Product of zeroes = c/a

$\sf \dfrac{4}{3} \times \dfrac{-3}{2} = \dfrac{-2}{1}$

$\sf\dfrac{-12}{6} = -2$

$\sf -2 = -2$