if alpha and beeta are the zero of the polynomial p(x)=3x2--12x++15 fi

1.	if alpha and beeta are the zero of the polynomial p(x)=3x2--12x++15 find the value of alpha square ++beeta square
Answer» Solution : $\bf{\green{\underline{\underline{\bf{Given\::}}}}}$ If α and β are the ZEROES of the polynomial p(x) = 3x² - 12X + 15. $\bf{\green{\underline{\underline{\bf{To\:find\::}}}}}$ The value of α² + β². $\bf{\green{\underline{\underline{\bf{Explanation\::}}}}}$ We have quadratic polynomial as we compared with ax² + bx + c = 0; a = 3 b = -12 c = 15 $\bf{\orange{\underline{\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}}}}$ $\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:(x)^{2} }{Coefficient\:of\:x} }\\\\\\\mapsto\sf{\alpha +\beta =\dfrac{-(-12)}{3} }\\\\\\\mapsto\sf{\alpha +\beta =\cancel{\dfrac{12}{3}} }\\\\\\\mapsto\sf{\red{\alpha +\beta =4}}$ $\bf{\orange{\underline{\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}}}}$ $\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:x} }\\\\\\\mapsto\sf{\alpha \times \beta =\dfrac{15}{3} }\\\\\\\mapsto\sf{\alpha \times \beta =\cancel{\dfrac{15}{3}} }\\\\\\\mapsto\sf{\red{\alpha \times \beta =5}}$ Now; $\mapsto\sf{\alpha^{2} +\beta ^{2} =(\alpha +\beta)^{2} -2\alpha \beta }\\\\\mapsto\sf{\alpha^{2} +\beta^{2} =(4 )^{2} -2(5) }\\\\\mapsto\sf{\alpha^{2} +\beta^{2} =16 -10}\\\\\mapsto\alpha^{2} +\beta ^{2}=16-10}\\ \\\mapsto\sf{\orange{\alpha^{2}+ \beta ^{2} =6}}$

if alpha and beeta are the zero of the polynomial p(x)=3x2--12x++15 find the value of alpha square ++beeta square

Answer»

Solution :

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

If α and β are the ZEROES of the polynomial p(x) = 3x² - 12X + 15.

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

The value of α² + β².

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

We have quadratic polynomial as we compared with ax² + bx + c = 0;

a = 3
b = -12
c = 15

$\bf{\orange{\underline{\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:(x)^{2} }{Coefficient\:of\:x} }\\\\\\\mapsto\sf{\alpha +\beta =\dfrac{-(-12)}{3} }\\\\\\\mapsto\sf{\alpha +\beta =\cancel{\dfrac{12}{3}} }\\\\\\\mapsto\sf{\red{\alpha +\beta =4}}$

$\bf{\orange{\underline{\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:x} }\\\\\\\mapsto\sf{\alpha \times \beta =\dfrac{15}{3} }\\\\\\\mapsto\sf{\alpha \times \beta =\cancel{\dfrac{15}{3}} }\\\\\\\mapsto\sf{\red{\alpha \times \beta =5}}$

Now;

$\mapsto\sf{\alpha^{2} +\beta ^{2} =(\alpha +\beta)^{2} -2\alpha \beta }\\\\\mapsto\sf{\alpha^{2} +\beta^{2} =(4 )^{2} -2(5) }\\\\\mapsto\sf{\alpha^{2} +\beta^{2} =16 -10}\\\\\mapsto\alpha^{2} +\beta ^{2}=16-10}\\ \\\mapsto\sf{\orange{\alpha^{2}+ \beta ^{2} =6}}$