Question

if alpha and beta are the roots of x^2 + bx + c =0, then the value of Alpha^2/ beta^2 +beta^2 /Alpha^2 is....

Please solve it... it's urgent​

Answer 1

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

$\sf{The \ value \ of \ \frac{\alpha^{2}}{\beta^{2}}+\frac{\beta^{2}}{\alpha^{2}} \ is \ \frac{b^{4}-4b^{2}c+2c^{2}}{c^{2}}}$

$\sf\orange{}$

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

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

$\sf{\implies{Roots \ are \ \alpha \ and \ \beta}}$

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

$\sf{\frac{\alpha^{2}}{\beta^{2}}+\frac{\beta^{2}}{\alpha^{2}}}$

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

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

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

$\sf{Here, \ a=1, \ b=b \ and \ c=c}$

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

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

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

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

$\sf{\implies{\frac{\alpha^{2}}{\beta^{2}}+\frac{\beta^{2}}{\alpha^{2}}}}$

$\sf{\implies{\frac{\alpha^{4}}{\alpha^{2}×\beta^{2}}+\frac{\beta^{4}}{\alpha^{2}×\beta^{2}}}}$

$\sf{\implies{\frac{\alpha^{4}+\beta^{4}}{(\alpha×\beta)^{2}}}}$

$\sf{By \ identity}$

$\sf{a^{4}+b^{4}=(a+b)^{4}-4ab(a^{2}+b^{2})-6(ab)^{2}}$

$\sf{\therefore{a^{4}+b^{4}=(a+b)^{4}-4ab[(a+b)^{2}-2ab]-6(ab)^{2}}}$

$\sf{\implies{\frac{(-b)^{4}-4(c)[(-b)^{2}-2c]-6c^{2}}{c^{2}}}}$

$\sf{\implies{\frac{b^{4}-4b^{2}c+8c^{2}-6c^{2}}{c^{2}}}}$

$\sf{\implies{\frac{b^{4}-4b^{2}c+2c^{2}}{c^{2}}}}$

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