Question

if alpha and beta are the two zeros of the polynomial 3 x square - 7 x minus 6 then find a polynomial whose zeros are 2 alpha + 3 beta and 3 alpha + 2 beta​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

If α and β are the two zeroes of the polynomial 3x² - 7x - 6.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

A polynomial whose zeroes are 2α + 3β and 3α + 2β.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have p(x) = 3x² -7x - 6

Zero of the polynomial is p(x) = 0

So;

$\leadsto\sf{3x^{2} -7x-6=0}\\\\\leadsto\sf{3x^{2} -9x+2x-6=0}\\\\\leadsto\sf{3x(x-3)+2(x-3)=0}\\\\\leadsto\sf{(x-3)(3x+2)=0}\\\\\leadsto\sf{x-3=0\:\:Or\:\:x=3x+2=0}\\\\\leadsto\sf{x=3\:\:Or\:\:3x=-2}\\\\\leadsto\sf{\pink{x=3\:\:Or\:\:x=-\dfrac{2}{3} }}$

∴ The α = 3 and β = -2/3 are the zeroes of the polynomial.

$\bf{\underline{\underline{\tt{A.T.Q\::}}}}}$

• $\sf{\alpha _{r}=(2\alpha +3\beta )}$
• $\sf{\beta _{r}=(3\alpha +2\beta )}$

Putting the value of the zeroes in this polynomial :

$\bf{\green{\underline{\underline{\tt{Sum\:of\:the\:zeroes\::}}}}}$

$\mapsto\sf{\alpha_{r} +\beta_{r} =\dfrac{Coefficient\:of\:(x)^{2} }{Coefficient\:of\:(x)} }\\\\\\\mapsto\sf{(2\alpha +3\beta )+(3\alpha +2\beta )}\\\\\\\mapsto\sf{2\alpha +3\alpha +3\beta +2\beta }\\\\\\\mapsto\sf{5\alpha +5\beta }\\\\\\\mapsto\sf{5(\alpha +\beta )}\\\\\\\mapsto\sf{5\times \bigg[3+ \bigg(-\dfrac{2}{3} \bigg)\bigg]}\\\\\\\mapsto\sf{5\times \bigg[3-\dfrac{2}{3} \bigg]}\\\\\\\mapsto\sf{5\times \bigg[\dfrac{9-2}{3} \bigg]}\\\\\\\mapsto\sf{5\times \dfrac{7}{3} }$

$\mapsto\sf{\pink{\dfrac{35}{3} }}$

$\bf{\green{\underline{\underline{\tt{Product\:of\:the\:zeroes\::}}}}}$

$\mapsto\sf{\alpha_{r} \times \beta_{r} =\dfrac{Constant\:term}{Coefficient\:of\:x} }\\\\\\\mapsto\sf{(2\alpha +3\beta )\times (3\alpha +2\beta )}\\\\\\\mapsto\sf{2\alpha (3\alpha +2\beta )+3\beta (3\alpha +2\beta )}\\\\\\\mapsto\sf{6\alpha ^{2} +4\alpha \beta +9\alpha \beta +6\beta ^{2} }\\\\\\\mapsto\sf{6\alpha ^{2} +6\beta ^{2} +13\alpha \beta }\\\\\\\mapsto\sf{6(\alpha ^{2} +\beta ^{2} )+13\alpha \beta }\\\\\\\mapsto\sf{6[(\alpha +\beta )^{2} -2\alpha \beta] +13\alpha \beta }$

$\mapsto\sf{6\bigg[\bigg(\dfrac{7}{3} \bigg)^{2} -2\times(-2)\bigg]+13(-2)}\\\\\\\mapsto\sf{6\bigg[\dfrac{49}{9} +4\bigg]+(-26)}\\\\\\\mapsto\sf{6\bigg[\dfrac{49+36}{9} \bigg]-26}\\\\\\\mapsto\sf{6\bigg[\dfrac{85}{9} \bigg]-26}\\\\\\\mapsto\sf{\cancel{6}\times \dfrac{85}{\cancel{9}} -26}\\\\\\\mapsto\sf{\dfrac{170}{3} -26}\\\\\\\mapsto\sf{\dfrac{170-78}{3}} \\\\\\\mapsto\sf{\pink{\dfrac{92}{3}}}$

Now;

The polynomial is required :

$\mapsto\sf{x^{2} -(Sum\:of\:zeroes)+(Product\:of\:zeroes)}\\\\\\\mapsto\sf{x^{2} -\dfrac{35}{3}x +\dfrac{92}{3}}$

Answer 2

$\tt{\huge{\orange {------------- }}}$

QUESTION :

if alpha and beta are the two zeros of the polynomial 3 x square - 7 x minus 6 then find a polynomial whose zeros are 2 alpha + 3 beta and 3 alpha + 2 beta

SOLUTION :

Given Polynomial :

3x^2 - 7x - 6 = 0

=> Factorising :

3x^2 - 9 x + 2x - 6 = 0

=> 3x ( x - 3 ) + 2 ( x - 3 ) = 0

=> ( 3x + 2 ) ( x - 3 ) = 0

=> Alpha = -2 / 3

=> Beta = 3

Sum of Zeroes Of New Polynomial : 5 Alpha + 5 Beta = 35 / 3

Similarly :

Product of Zeroes = -92 /.3

New Polynomial :

X^2 - ( Sum of Zeroes ) + ( Product of Zeroes )

=> X^2 - 35 / 3 X - 92 / 3.....[ A ]