Question

if alpa and beta are the zeros of the quadratic polymial f(x)=ax^2+bx+c, then evaluat alpa-beta​

Answer 1

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:}}}}$

$\sf{The \ value \ of \ \alpha-\beta \ is \ \frac{\sqrt{b^{2}-4ac}}{a}}$

$\sf\orange{Given:}$

$\sf{The \ given \ quadratic \ polynomial \ is}$

$\sf{\implies{f(x)=ax^{2}+bx+c}}$

$\sf{\implies{\alpha \ and \ \beta \ are \ the \ zeroes.}}$

$\sf\pink{To \ find:}$

$\sf{The \ value \ of \ \alpha-\beta}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ quadratic \ polynomial \ is}$

$\sf{\implies{f(x)=ax^{2}+bx+c}}$

$\sf{Sum \ of \ zeroes=\frac{-b}{a}}$

$\sf{\implies{\therefore{\alpha+\beta=\frac{-b}{a}...(1)}}}$

$\sf{Product \ of \ zeroes=\frac{c}{a}}$

$\sf{\implies{\therefore{\alpha\beta=\frac{c}{a}...(2)}}}$

$\sf{According \ to \ identity}$

$\sf{(a-b)^{2}=(a+b)^{2}-4ab}$

$\sf{\implies{(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta}}$

$\sf{...from \ (1) \ and (2)}$

$\sf{\implies{(\alpha-\beta)^{2}=(\frac{-b}{a})^{2}-4\times\frac{c}{a}}}$

$\sf{\implies{(\alpha-\beta)^{2}=\frac{b^{2}}{a^{2}}-\frac{4c}{a}}}$

$\sf{\implies{(\alpha-\beta)^{2}=\frac{b^{2}-4ac}{a^{2}}}}$

$\sf{Taking \ square \ root \ of \ both \ sides}$

$\sf{\implies{\alpha-\beta=\frac{\sqrt{b^{2}-4ac}}{a}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ \alpha-\beta \ is \ \frac{\sqrt{b^{2}-4ac}}{a}}}}$