Question

find the zeroes of the following polynomial and verify their relationship between the zeroes x² + 3x + 2​

Answer 1

☯GIVEN :

• $\sf {x^2 + 3x + 2}$

☯TO DO :

• Find the zeros and verify the relation between zeros and coefficients.

➲SOLUTION :

$:\implies \sf{x^2 + 3x + 2 = 0}$

$:\implies \sf{x^2 +x + 2x + 2 = 0}$

$:\implies \sf{x \left(x + 1 \right)+ 2 \left(x + 1 \right) = 0}$

$:\implies \sf{\left(x + 1\right) \left(x + 2 \right) = 0}$

$:\implies \sf{\left(x +1 \right) = 0 \: , \: \left(x + 2 \right) = 0}$

$:\implies\sf{x = 0 - 1\: , \: x = 0 - 2}$

$\mapsto \underline {\huge{\boxed{ \purple{\bf{x = - 1\: , \: x = - 2}}}}}$

★Verifying the relation between zeros and coefficients :

Here,

• a = 1

• b = 3

• c = 2

▪Sum of zeros :

$:\leadsto \sf {\alpha + \beta = \dfrac{ - b}{a} }$

$:\leadsto \sf {-1+(-2) = \dfrac{ - 3}{1} }$

$:\leadsto \sf {-1-2 = -3 }$

$:\leadsto {\huge{\boxed{\bf{ \pink{-3= -3}}}}}$

▪Product of zeros :

$:\leadsto \sf{\alpha \beta = \dfrac{c}{ a}}$

$:\leadsto \sf{-1×(-2) = \dfrac{ 2}{ 1}}$

$:\leadsto{\underline {\huge{\boxed{ \blue{\bf{2= 2}}}}}}$

$\huge {\green{\therefore }}$ Verified.