Math, asked by Gunnik36, 10 months ago

if alpha and beta are the zeroes of the quadratic polynomial fx =3x^2 -7x -6 ,find a polynomial whose zeroes are 2 alpha +3 beta(3 alpha +2 beta)

Answers

Answered by TheProphet
1

Solution :

We have quadratic polynomial p(x) = 3x² - 7x - 6

Zero of the polynomial p(x) = 0

\longrightarrow\sf{3x^{2} -7x-6=0}\\\\\longrightarrow\sf{3x^{2} -9x+2x-6=0}\\\\\longrightarrow\sf{3x(x-3)+2(x-3)=0}\\\\\longrightarrow\sf{(x-3)(3x+2)=0}\\\\\longrightarrow\sf{x-3=0\:\:\:Or\:\:\:3x+2=0}\\\\\longrightarrow\sf{x=3\:\:\:Or\:\:\:3x=-2}\\\\\longrightarrow\bf{x=3\:\:\:Or\:\:\:x=-2/3}

∴ α = 3 & β = -2/3 are two zeroes of the polynomial.

Now;

\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}}

\mapsto\sf{\alpha +\beta }\\\\\mapsto\sf{3+\bigg(-\dfrac{2}{3} \bigg)}\\\\\\\mapsto\sf{3-\dfrac{2}{3} }\\\\\\\mapsto\sf{\dfrac{9-2}{3} }\\\\\\\mapsto\bf{\dfrac{7}{3}..................(1) }

\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}}

\mapsto\sf{\alpha \times \beta }\\\\\mapsto\sf{3\times \bigg(-\dfrac{2}{3} \bigg)}\\\\\\\mapsto\bf{\dfrac{-6}{3} ..............(2)}

\underbrace{\bf{According\:to\:the\:question\::}}}}

Zero of the polynomial α & β = 2α + 3β & 3α + 2β

\underline{\mathcal{SUM\:OF\:ZEROES\::}}}

\mapsto\sf{\alpha +\beta }\\\\\mapsto\sf{(2\alpha + 3\beta ) + (3\alpha + 2\beta )}\\\\\mapsto\sf{2\alpha +3\beta + 3\alpha + 2\beta }\\\\\mapsto\sf{2\alpha +3\alpha +3\beta +2\beta }\\\\\mapsto\sf{5\alpha +5\beta }\\\\\mapsto\sf{5(\alpha +\beta )}\\\\\mapsto\sf{5\times 7/3}\\\\\mapsto\bf{35/3}

\underline{\mathcal{PRODUCT\:OF\:ZEROES\::}}}

\mapsto\sf{\alpha \beta }\\\\\mapsto\sf{(2\alpha +3\beta )(3\alpha +2\beta )}\\\\\mapsto\sf{6\alpha ^{2} +4\alpha \beta +9\alpha \beta +6\beta ^{2} }\\\\\mapsto\sf{6\alpha ^{2} +6\beta ^{2}+13\alpha\beta}\\\\\mapsto\sf{6(\alpha^{2} +\beta^{2} )+13\alpha \beta}\\\\\mapsto\sf{6[(\alpha+\beta)^{2} -2\alpha \beta]+13\alpha \beta}\\\\\mapsto\sf{6\bigg[\bigg(\dfrac{7}{3} \bigg)^{2} -2\bigg(-\dfrac{6}{3} \bigg)\bigg]+13\times \bigg(-\dfrac{6}{3} \bigg)\:\:[\therefore from(1)\:\:\&\:\:(2)]}\\\\\\

\mapsto\sf{6\bigg[\bigg(\dfrac{49}{9} \bigg)\bigg]-\bigg(\dfrac{-12}{3} \bigg)\bigg]-\dfrac{78}{3}}\\\\\\\mapsto\sf{6\bigg[\dfrac{49}{9} +\dfrac{12}{3} \bigg]-\dfrac{78}{3} }\\\\\\\mapsto\sf{6\bigg[\dfrac{49+36}{9} \bigg]-\dfrac{78}{3} }\\\\\\\mapsto\sf{6\bigg[\dfrac{85}{9} \bigg]-\dfrac{78}{3} }\\\\\\\mapsto\sf{\dfrac{510}{9} -\dfrac{78}{3} }\\\\\\\mapsto\sf{\dfrac{510-234}{9} }\\\\\mapsto\bf{276/9}

Thus;

\boxed{\bf{The\:required\:quadratic\:polynomial\::}}}}

\longrightarrow\sf{x^{2} -(sum\:of\:zeroes)x+(product\:of\:zeroes)}\\\\\\\longrightarrow\sf{x^{2} -\bigg(\dfrac{35}{3} \bigg)x + \bigg(\dfrac{276}{9} \bigg)}

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