Question

write sum of zeroes an product of zeroes of the following polynomial 5x²-x-4​

Answer 1

Question:-

$\sf{Write\:the\:sum\:and\:product\:of\:zeroes\:of\:the\:polynomial\:5x^2-x-4}$

Solution:-

$\sf{p(x) = 5x^2 - x - 4}$

= $\sf{By\:Splitting\:the\:middle\:term}$

= $\sf{5x^2 - 5x + 4x - 4 = 0}$

= $\sf{5x(x-1) + 4(x-1) = 0}$

= $\sf{(x-1)(5x+4) = 0}$

Either,

= $\sf{x - 1 = 0}$

=> $\sf{x = 1}$

Or,

= $\sf{5x + 4 = 0}$

=> $\sf{5x = -4}$

=> $\sf{x = \dfrac{-4}{5}}$

$\sf{\therefore The\: zeroes\: of \:the\:polynomial\:5x^2-x-4\:are\:1 \:and \: \dfrac{-4}{5}}$

$\sf{Let\: \alpha = 1\: and \: \beta = \dfrac{-4}{5}}$

Therefore,

$\sf{Product\:of\:Zeroes = \alpha\beta = 1\times \dfrac{-4}{5} = \dfrac{-4}{5}}$

$\sf{Sum\:of\:Zeroes = \alpha + \beta = 1 + \dfrac{-4}{5}}$

= $\sf{\dfrac{5-4}{5} = \dfrac{1}{5}}$

Verification:-

$\sf{Sum\:of\:zeroes = \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

= $\sf{\dfrac{1}{5} = \dfrac{1}{5}\:[Verified]}$

$\sf{Product\:of\:zeroes = \dfrac{Constant\:term}{Coefficient\:of\:x^2}}$

= $\sf{\dfrac{-4}{5} = \dfrac{-4}{5}\: [Verified]}$