Question

Q2.If the sum of the zeroes of the polynomial p(x) = kx' +2x+3k is equal to their product, then
find two value of 'k'.​ .​

Answer 1

$given \: polynomial = kx2−2x−3k </p><p></p><p>$

$if ax2+bx+c is \: a \: \\ polynomial \: then \: its \: \\ sum \: of \: roots \: is \: given \: \\ by \:  -b/a \: and \: product \: of \\ roots \: is \: given \: by \:  c/a$

$∴ \:  sum \: of \: roots \: of \\ given \: polynomial  \: \\ =−(−2)/k=2/k</p><p></p><p></p><p>$

$product \: of \: roots  \: \\ =−3k/k=−3$

$it \: is \: given \: that \: \\ sum \: of \: roots \: \\  = product \: of \: roots </p><p></p><p></p><p>$

$\frac{2}{k} = 3$

$k = \frac{ - 2}{3}$

$hence \: verified$

Answer 2

$\huge \sf {\purple{\underline {\pink {\underline { Answer᭄\ }}}}}$

Kx² + 2x + 3k

a = k , b = 2 , c = 3k

Finding sum of the zeroes ,

α + β = - b /a

α + β = - 2 / k

Now , Finding the product of the zeroes,

αβ = c / a

αβ = 3k / k = 3

Given that :-

Sum of the zeroes = Product of the zeroes

i.e , α + β = αβ

- 2 / k = 3

- 2 = 3k

$\sf\red{↠}$k = - 2 / 3

${\huge{\underline{\small{\mathbb{\red{HOPE\:IT\:HELPS\:UH :)}}}}}}$