Question

Find the zeroes of the following quardratic and verify the relationship between the zeroes and the coefficient
6x square-3-7x

Answer 1

AnswEr :

$\underline{\bigstar\:\textsf{Let's Head to the Question :}}$

$:\implies\tt 6x^2-3-7x=0\\\\\\:\implies\tt 6x^2-7x-3=0\\\\\\:\implies\tt6x^2-(9-2)x-3=0\\\\\\:\implies\tt6x^2-9x+2x-3=0\\\\\\:\implies\tt3x(2x-3)+1(2x-3)=0\\\\\\:\implies\tt(3x+1)(2x-3)=0\\\\\\:\implies\underline{\boxed{\tt \alpha =\dfrac{ -1}{3} \quad or \quad \beta =\dfrac{3}{2}}}$

$\rule{200}{1}$

$\underline{\bigstar\:\textsf{Sum of Zeroes :}}$

$\twoheadrightarrow\:\:\:\tt Sum\:of\:Zeroes=\dfrac{- \:b}{a}\\\\\\\twoheadrightarrow\:\:\:\tt \alpha + \beta =\dfrac{- \:b}{a}\\\\\\\twoheadrightarrow\:\:\:\tt \dfrac{ - 1}{3} + \dfrac{3}{2} = \dfrac{-\:( - 7)}{6}\\\\\\\twoheadrightarrow\:\:\:\tt\dfrac{ -1(2) + 3(3)}{6} = \dfrac{7}{6}\\\\\\\twoheadrightarrow\:\:\:\tt \dfrac{ - 2 + 9}{6} = \dfrac{7}{6}\\\\\\\twoheadrightarrow\:\:\: \large\underline{\boxed{\orange{\tt\dfrac{7}{6} = \dfrac{7}{6}}}}$

$\rule{150}{2}$

$\underline{\bigstar\:\textsf{Product of Zeroes :}}$

$\twoheadrightarrow\:\:\:\tt Product\:of\:Zeroes=\dfrac{c}{a} \\\\\\\twoheadrightarrow\:\:\:\tt \alpha \times\beta = \dfrac{c}{a}\\\\\\\twoheadrightarrow\:\:\:\tt \dfrac{ - 1}{3} \times\dfrac{3}{2} = \dfrac{ - 3}{6}\\\\\\\twoheadrightarrow\:\:\: \large\underline{\boxed{\orange{\tt\dfrac{ - \: 1}{2} = \dfrac{ -\:1}{2}}}}$

Answer 2

$\bf{\Large{\boxed{\sf{ANSWER\::}}}}}$

$\bf{\large{\underline{\mathfrak{Given\::}}}}$

Quadratic polynomial are 6x² - 3 - 7x.

$\bf{\large{\underline{\mathfrak{To\:find\::}}}}$

The zeroes and verify the relationship between the zeroes.

$\bf{\Large{\underline{\underline{\tt{\green{Explanation\::}}}}}}$

Let p(x) = 0

$\bf{\large{\underline{\underline{\tt{\red{A.T.Q\::}}}}}}$

$\dashrightarrow\tt{6x^{2} -3-7x}\\\\\\\dashrightarrow\tt{6x^{2} -7x-3=0}\\\\\\\dashrightarrow\tt{6x^{2} -9x+2x-3=0}\\\\\\\dashrightarrow\tt{3x(2x-3)+1(2x-3)=0}\\\\\\\dashrightarrow\tt{(2x-3)(3x+1)=0}\\\\\\\dashrightarrow\tt{2x-3=0\:\:\:\:\:\:Or\:\:\:\:\:\:3x+1=0}\\\\\\\dashrightarrow\tt{2x=3\:\:\:\:\:\:\:Or\:\:\:\:\:\;3x=-1}\\\\\\\dashrightarrow\tt{\blue{x=\dfrac{3}{2} \:\:\:\:Or\:\:\:\:\:x=-\dfrac{1}{3} }}$

∴We have get the two zeroes α = 3/2 & β = -1/3.

Now,

$\bf{\large{\underline{\underline{\sf{Verify\:the\:relationship\:between\:the\:zeroes\:and\:coefficient\:of\:polynomials.}}}}}}}$

We have;

• a = 6
• b = -7
• c = -3

$\bf{\large{\underline{\tt{\red{\blacksquare{Sum\;of\:zeroes\::}}}}}}$

$\Rightarrow\tt{\alpha +\beta =-\dfrac{b}{a} \:\:\ \longrightarrow\:\:\:\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\\\Rightarrow\tt{\dfrac{3}{2} +\big(-\dfrac{1}{3} \big)=\dfrac{-(-7)}{6} }\\\\\\\\\Rightarrow\tt{\dfrac{3}{2} -\dfrac{1}{3} =\dfrac{7}{6} }\\\\\\\\\Rightarrow\tt{\dfrac{9-2}{6} =\dfrac{7}{6} }\\\\\\\\\Rightarrow\tt{\green{\dfrac{7}{6} =\dfrac{7}{6}\: \longrightarrow\dfrac{coefficient\:of\:x}{coefficient\:of\:x^{2} } }}$

$\bf{\large{\underline{\tt{\red{\blacksquare{Product\;of\:zeroes\::}}}}}}$

$\Rightarrow\tt{\alpha \times \beta =\dfrac{c}{a} \:\:\: \longrightarrow\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\\\Rightarrow\tt{\dfrac{3}{2} \times -\dfrac{1}{3} =\dfrac{-3}{6}}\\\\\\\\\Rightarrow\tt{\green{\dfrac{-3}{6} =\dfrac{-3}{6} \:\:\:\:\: \longrightarrow\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }}$