Question

if p(x) =ax2 + bx +c is a quadratic polynomial, then what is the relation of -b/a with zeros of p(x)?​

Answer 1

Let $\displaystyle\sf {x_1}$ and $\displaystyle\sf {x_2}$ be the zeroes of the equation $\displaystyle\sf {ax^2+bx+c=0.}$

Let $\displaystyle\sf {k}$ be a non - zero constant. Then we have,

$\displaystyle\sf{\longrightarrow k(x-x_1)(x-x_2)=ax^2+bx+c}$

$\displaystyle\sf{\longrightarrow k(x^2-x_1x-x_2x+x_1x_2)=ax^2+bx+c}$

$\displaystyle\sf{\longrightarrow k(x^2-(x_1+x_2)x+x_1x_2)=ax^2+bx+c}$

$\displaystyle\sf{\longrightarrow kx^2-k(x_1+x_2)x+kx_1x_2=ax^2+bx+c}$

Equating corresponding coefficients,

$\displaystyle\sf{\longrightarrow k=a\quad\quad\dots(1)}$

and,

$\displaystyle\sf{\longrightarrow -k(x_1+x_2)=b}$

From (1),

$\displaystyle\sf{\longrightarrow -a(x_1+x_2)=b}$

$\displaystyle\sf {\longrightarrow\underline {\underline {x_1+x_2=-\dfrac {b}{a}}}}$

That is, the sum of zeroes is equal to the ratio $\displaystyle\sf {-\dfrac {b}{a}.}$ This is the relation.

Also,

$\displaystyle\sf{\longrightarrow kx_1x_2=c}$

From (1),

$\displaystyle\sf{\longrightarrow ax_1x_2=c}$

$\displaystyle\sf{\longrightarrow x_1x_2=\dfrac {c}{a}}$

We get the product of roots is equal to the ratio $\displaystyle\sf {\dfrac {c}{a}}$ too.