Question

find a quadratic polynomial whose zeroes are -4 and 3 verify the relationship between the zeroes and the coefficents

Answer 1

S O L U T I O N :

Let the zeroes of the polynomial be α & β which are given α = -4 & β = 3 respectively.

$\underline{\underline{\tt{According\:to\:the\:question\::}}}$

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

$\longrightarrow\sf{\alpha + \beta }$

$\longrightarrow\sf{-4 + 3 }$

$\longrightarrow\bf{-1}$

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

$\longrightarrow\sf{\alpha \times \beta }$

$\longrightarrow\sf{-4 \times 3 }$

$\longrightarrow\bf{-12}$

Now,the required polynomial are;

$\mapsto\tt{x^{2} - (sum\:of\:zereos)x + (product\:of\:zeroes)}$

$\mapsto\tt{x^{2} - (-1)x + (-12)}$

$\mapsto\bf{x^{2} +1x -12}$

V E R I F I C A T I O N :

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

• a = 1
• b = 1
• c = -12

Now,

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

$\mapsto\tt{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}}\bigg\rgroup}$

$\mapsto\tt{-4 + 3 =\dfrac{-1}{1} }$

$\mapsto\bf{-1 = -1}$

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

$\mapsto\tt{\alpha \times \beta =\dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2}}\bigg\rgroup}$

$\mapsto\tt{-4 \times 3 =\dfrac{-12}{1} }$

$\mapsto\bf{-12 = -12}$

Thus,

The relationship between zeroes & coefficient are verified .