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

$\large \orange{\underline{ \orange{\underline{\bold{\purple{\overbrace{\pink{ \underbrace{ |\:\:\:\: \rm{ \mathfrak{ \red{ \huge{answer}}} \:\:\:\:|} }}}}}}}}}$

$\rm{ \bold{ \green{\underline{ \pink{ \underline{ \red{ \underbrace{ \overbrace{\blue{\mid \:\:\:\:\:\: k= \dfrac{-2}{3}\:\:\:\:\:\: \mid}}}}}}}}}}$

$\rm{ \green{ \underline{ \red{ \overbrace{ \pink{ \mathfrak{ \:\:\:explanation\:in\:details \:\: \downarrow}}}}}}}$

$\large\underline\bold{ \mathcal{ \pink{GIVEN:}}}$

$\mapsto quadratic\:equation:- \blue{ kx^2+2x+3k=0 } \\ \\ \mapsto sum \:of\:roots= \orange{(\alpha +\beta ) } \\ \\ \mapsto product\:of\:roots :- \green{(\alpha \beta )}$

$\large\underline\bold{ \mathcal{ \pink{To\:Find:}}}$

$\rm{ \red{ \star}} \: the\:value\:of\:K .$

$\large\underline\bold{ \underline{ \mathfrak{ \purple{Solving:}}}}$

$\leadsto a= k \\ \leadsto b= 2 \\ \leadsto c= 3 \\ \\ \rm{ \green{\circ }} \: \red{ sum \:of\:zeroes= \dfrac{-b}{a}} \\ \\ :\implies \underline{\underline{(\alpha + \beta )= \pink{ \dfrac{-2}{k}}} } \\ \\ \therefore Now \:for\: product \\ \\ \rm{\green{\circ}} \:\red{ product \:of\:zeroes= \dfrac{c}{a}} \\ \\ :\implies \dfrac{ 3\:\cancel{k\:}}{ \cancel{k\:}} \\ \\ :\implies \underline{ \underline{ ( \alpha \beta ) = \pink{3 }}} \\ \\ \green{ as\:given\:sum\:of\:zeroes \:= \:. product\:of\:zeroes.} \\ \\ \rightarrow (\alpha \beta) = (\alpha +\beta) \\ \\ :\implies 3= \dfrac{-2}{k} \\ \\ :\implies 3k= -2 \\ \\ :\implies k= \dfrac{-2}{3} \\ \\ \rm{ \pink{\overbrace{ \overline{ \red{ \underbrace{\underline{ \blue{ \mid\:\:\:\:\:\:\:\:\:\:\:k=\dfrac{-2}{3} \:\:\:\:\:\:\:\:\:\:\: \mid}}}}}}}}$

$\green{ \leftarrow} ^{\large{ \red{\uparrow}}} _{ \large{\pink{\downarrow}}} \blue{\rightarrow}$

$\rm{ \bold{\red{ \underline{ \purple{ \underline{ \overline{ \purple{ \overline{ \green{ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}}}}}}}}}$

Answer 2

Step-by-step explanation:

quadraticequation:−kx

2

+2x+3k=0

↦sumofroots=(α+β)

↦productofroots:−(αβ)

\begin{gathered}\large\underline\bold{ \mathcal{ \pink{To\:Find:}}} \\ \end{gathered}

ToFind:

\rm{ \red{ \star}} \: the\:value\:of\:K .⋆thevalueofK.

\begin{gathered}\large\underline\bold{ \underline{ \mathfrak{ \purple{Solving:}}}}\\ \\\end{gathered}

Solving:

\begin{gathered}\leadsto a= k \\ \leadsto b= 2 \\ \leadsto c= 3 \\ \\ \rm{ \green{\circ }} \: \red{ sum \:of\:zeroes= \dfrac{-b}{a}} \\ \\ :\implies \underline{\underline{(\alpha + \beta )= \pink{ \dfrac{-2}{k}}} } \\ \\ \therefore Now \:for\: product \\ \\ \rm{\green{\circ}} \:\red{ product \:of\:zeroes= \dfrac{c}{a}} \\ \\ :\implies \dfrac{ 3\:\cancel{k\:}}{ \cancel{k\:}} \\ \\ :\implies \underline{ \underline{ ( \alpha \beta ) = \pink{3 }}} \\ \\ \green{ as\:given\:sum\:of\:zeroes \:= \:. product\:of\:zeroes.} \\ \\ \rightarrow (\alpha \beta) = (\alpha +\beta) \\ \\ :\implies 3= \dfrac{-2}{k} \\ \\ :\implies 3k= -2 \\ \\ :\implies k= \dfrac{-2}{3} \\ \\ \rm{ \pink{\overbrace{ \overline{ \red{ \underbrace{\underline{ \blue{ \mid\:\:\:\:\:\:\:\:\:\:\:k=\dfrac{-2}{3} \:\:\:\:\:\:\:\:\:\:\: \mid}}}}}}}} \end{gathered}

⇝a=k