Question

find the zeros of the quadratic polynomial 2 x ^2 - 8 x + 6 and verify the relation between zeros and its coefficient

Answer 1

AnswEr :

$\bf{\green{\underline{\underline{\bf{Given\::}}}}}$

The quadratic polynomial 2x² - 8x + 6 and verify the relationship between the zeroes and it's coefficient.

$\bf{\green{\underline{\underline{\bf{To\:find\::}}}}}$

The zeroes.

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

We have, quadratic polynomial = 2x² - 8x + 6;

$\leadsto\tt{2x^{2} -8x+6=0}\\\\\leadsto\tt{2x^{2} -6x-2x+6=0}\\\\\leadsto\tt{2x(x-3)-2(x-3)=0}\\\\\leadsto\tt{(x-3)(2x-2)=0}\\\\\leadsto\tt{(x-3)=0\:\:\:\:Or\:\:\:\:(2x-2)=0}\\\\\leadsto\tt{x-3=0\:\:\:\:\:Or\:\:\:\:\:2x-2=0}\\\\\leadsto\tt{x=3\:\:\:\:Or\:\:\:\:2x=2}\\\\\leadsto\tt{x=3\:\:\:\:\:Or\:\:\:\:\:x=\cancel{\dfrac{2}{2} }}\\\\\leadsto\tt{\orange{x=3\:\:\:Or\:\:\:x=1}}$

Thus,

$\bf{\large{\underline{\sf{\red{We\:have\:zeroes\:are\:\alpha =3\:\:\&\:\:\beta =1.}}}}}$

∴ Relation between the zeroes and it's coefficient :

From quadratic polynomial :

• a = 2
• b = -8
• c = 6

$\bf{\large{\purple{\underline{\underline{\blacksquare{\bf{Sum\:of\:the\:zeroes\::}}}}}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:(x)^{2} }{Coefficient\:of\:(x)} }\\\\\\\\\mapsto\sf{3+1=4=\dfrac{-b}{a} }}$

$\bf{\large{\purple{\underline{\underline{\blacksquare{\bf{Product\:of\:zeroes\::}}}}}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:(x)} }\\\\\\\\\mapsto\sf{3\times 1=3=\dfrac{c}{a} }}$

Thus,

$\bf{\small{\red{\underline{\sf{The\:relation\:between\:the\:zeroes\:and\:it's\:coefficient\:of\:polynomial\:is\:\:Verified.}}}}}$

Answer 2

$2 {x}^{2} - 8x + 6 = 0 \\ 2 {x}^{2} - 2x - 6x + 6 = 0 \\ 2x (x - 1) - 6(x - 1) = 0 \\ (x - 1)(2x - 6) = 0 \\ x = 1 \: or \: x = \dfrac{6}{2} = 3 \\ so \: \alpha = 1 \: and \: \beta = 3 \\\\\ from \: the \: quadratic \: equation \\ \dfrac{ - b}{a} = \dfrac{ - ( - 8)}{2} = 4 (i) \\ \alpha + \beta = 1 + 3 = 4(ii) \\ from \:( i )\: and \:( ii )\: \\ \alpha + \beta = \dfrac{ - b}{a} \\ \\ \\ \dfrac{c}{a} = \frac{6}{2} = 3(iii) \\ \alpha \beta = 1 \times 3 = 3(iv) \\ from \: (iii ) \: and \: (iv) \\ \alpha \beta = \dfrac{c}{a}$