Math, asked by BrainlyDevilX, 2 months ago

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'.​ .​

Answers

Answered by muskan686
0

\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{ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}}}}}}}}}

Answered by Naimeesya
0

Your answer is in the attachment.

hope it helps...☺

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