Question

verify the relationship between the zeroes and coefficient of x +11x+16​

Answer 1

Solution :

p(x) = x² + 11x + 18

Zero of the polynomial p(x) = 0

$\longrightarrow\sf{x^{2} +11x+18=0}\\\\\longrightarrow\sf{x^{2} +9x+2x+18=0}\\\\\longrightarrow\sf{x(x+9)+2(x+9)=0}\\\\\longrightarrow\sf{(x+9)(x+2)=0}\\\\\longrightarrow\sf{x+9=0\:\:\:Or\:\:\:x+2=0}\\\\\longrightarrow\bf{x=-9\:\:\:Or\:\:\:x=-2}$

∴ α = -9 & β = -2 are the zeroes of the polynomial.

As we know that given quadratic polynomial compared with ax² + bx + c;

• a = 1
• b = 11
• c = 18

Now;

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

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } \bigg\rgroup }\\\\\\\mapsto\sf{-9+(-2)=\dfrac{-11}{1} }\\\\\\\mapsto\sf{-9-2=-11}\\\\\mapsto\bf{-11=-11}$

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

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } \bigg\rgroup }\\\\\\\mapsto\sf{-9\times (-2)=\dfrac{18}{1} }\\\\\\\mapsto\sf{-(-18)=18}\\\\\mapsto\bf{18=18}$

Thus;

The zeroes & relationship between zeroes & coefficient is verified .

p(x) = 2x² + x + 6

Zero of the polynomial p(x) = 0

$\longrightarrow\sf{2x^{2} +x-6=0}\\\\\longrightarrow\sf{2x^{2} +4x-3x-6=0}\\\\\longrightarrow\sf{2x(x+2)-3(x+2)=0}\\\\\longrightarrow\sf{(x+2)(2x-3)=0}\\\\\longrightarrow\sf{x+2=0\:\:\:Or\:\:\:2x-3=0}\\\\\longrightarrow\sf{x=-2\:\:\:Or\:\:\:2x=3}\\\\\longrightarrow\bf{x=-2\:\:\:Or\:\:\:x=3/2}$

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

As we know that given polynomial compared with ax² + bx + c;

• a = 2
• b = 1
• c = -6

Now;

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

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } \bigg\rgroup }\\\\\\\mapsto\sf{-2+\bigg(\dfrac{3}{2} \bigg)=\dfrac{-1}{2} }\\\\\\\mapsto\sf{\dfrac{-4+3}{2} =\dfrac{-1}{2} }\\\\\mapsto\bf{-1/2=-1/2}$

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

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } \bigg\rgroup }\\\\\\\mapsto\sf{-2\times \bigg(\dfrac{3}{2} \bigg)=\dfrac{-6}{2} }\\\\\\\mapsto\sf{-\cancel{6/2}=-\cancel{6/2}}\\\\\mapsto\bf{-3=-3}$

The zeroes & relationship between zeroes & coefficient is verified .