Question

find the zeros of a polynomial X square + 1 by 6 x minus 2 and verify the relationship between the zeros and coefficients​

Answer 1

$\sf{f(x) = x^2 + {\dfrac{1}{6}} x - 2}$


→ $\sf{f(x) = {\dfrac{6x^2 + x - 12}{6}} }$


$\sf{To \ find \ zeroes, \ f(x) = 0}$


$\sf{\therefore \ {\dfrac{6x^2 + x - 12}{6}} = 0}$


→ $\sf{6x^2 + x - 12 = 0}$


$\sf{\underline{By \ Middle \ Term \ Factorisation}}$


→ $\sf{6x^2 + 9x - 8x - 12 = 0}$


→ $\sf{3x(2x + 3) - 4(2x + 3) = 0}$


→ $\sf{(3x - 4)(2x + 3) = 0}$


$\sf{\underline{Using \ Zero \ Product \ Rule}}$


→ $\sf{3x - 4 = 0 \ and \ 2x + 3 = 0}$


→ $\sf{x = {\dfrac{4}{3}} \ and \ x = {\dfrac{- 3}{2}} }$


$\sf{Let \ \alpha \ and \ \beta \ be \ the \ zeroes.}$


$\sf{\therefore \ \alpha = {\dfrac{4}{3}}, \ \beta = {\dfrac{- 3}{2}} }$


$\large\sf\red{\underline{\underline{Verification}}}$


$\sf{On \ comparing \ the \ above \ polynomial \ with}$

$\sf{ax^2 + bx + c = 0, \ we \ get}$


→ $\sf{a = 1, \ b = {\dfrac{1}{6}}, \ c = - 2}$


$\sf{Now,}$


$\sf{\underline{Sum \ of \ zeroes \ :}}$


→ $\sf{\alpha + \beta = {\dfrac{- b}{a}}}$


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


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


$\sf{\underline{Product \ of \ zeroes \ :}}$


→ $\sf{\alpha \beta = {\dfrac{c}{a}} }$


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


→ $\sf{ - 2 = - 2}$

Answer 2

Step-by-step explanation:

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