Question

If alpha and beta are the zeroes of the quadratic polynomial fx = ax^2 +bx + c , then evaluate :
a(alpha^2/beta+beta^2/alpha) + b (alpha / beta + beta / alpha )..

SOLVE PLZZ DOSTO

Answer 1

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

$\sf\orange{Given:}$

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

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

$\sf{a(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha})+b(\frac{\alpha}{\beta})}$

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

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

$\sf{\implies{ax^{2}+bx+c}}$

$\sf{Here, \ a=a, \ b=b \ and \ c=c}$

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

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

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

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

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

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

$\sf{\therefore{\alpha^{2}+\beta^{2}=\frac{b^{2}}{a^{2}}-\frac{2c}{a}}}$

$\sf{\therefore{\alpha^{2}+\beta^{2}=\frac{b^{2}-2ac}{a^{2}}...(3)}}$

$\sf{\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)}$

$\sf{\therefore{\alpha^{3}+\beta^{3}=(\frac{-b}{a})^{3}-3×\frac{c}{a}×\frac{-b}{a}}}$

$\sf{\alpha^{3}+\beta^{3}=\frac{-b^{3}}{a^{3}}+\frac{3bc}{a^{2}}}$

$\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{-b^{3}-3abc}{a^{3}}...(4)}}$

________________________________

$\sf{\implies{a(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha})+b(\frac{\alpha}{\beta}+\frac{\beta}{\alpha})}}$

$\sf{\implies{a(\frac{\alpha^{3}+\beta^{3}}{\alpha×\beta})+b(\frac{\alpha^{2}+\beta^{2}}{\alpha×\beta})}}$

$\sf{From \ (1), \ (2), \ (3) \ and (4)}$

$\sf{\implies{a(\frac{\frac{-b^{3}-3abc}{3a^{3}}}{\frac{c}{a}})+b(\frac{\frac{b^{2}-2ac}{a^{2}}}{\frac{c}{a}})}}$

$\sf{\implies{a(\frac{-b^{3}-3abc}{3a^{3}}×\frac{a}{c})+b(\frac{b^{2}-2ac}{a^{2}}×\frac{a}{c})}}$

$\sf{\implies{\frac{-b^{3}-3a^{2}bc}{3a^{2}c}-\frac{b^{3}-2ac}{ac}}}$

$\sf{}$

Answer 2

Answer:

Correct me if there is any mistake...

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Step-by-step explanation: