If ax^2 + bx + c = 0 had two roots p and q and p^2 + q^2 = p^3 + q^3 then show that b^3 − 2a^2 c + ab^2 = 3abc.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 205353 by MATHEMATICSAM last updated on 17/Mar/24

$If {a x}^{2} + b x + c = 0 had two roots p and q$ $and p^{2} + q^{2} = p^{3} + q^{3} then show that$ $b^{3} - 2 a^{2} c + {a b}^{2} = 3 a b c .$
	Answered by sniper237 last updated on 17/Mar/24

	$p^{2} + q^{2} = {(p + q)}^{2} - 2 p q = \frac{b^{2}}{a^{2}} - \frac{2 c}{a}$ $p^{3} + q^{3} = {(p + q)}^{3} - 3 p q (p + q) = - \frac{b^{3}}{a^{3}} + \frac{3 b c}{a^{2}}$ $\overset{\times a^{3}}{\Rightarrow} - b^{3} + 3 a b c = {a b}^{2} - 2 a c$
	Answered by A5T last updated on 17/Mar/24

	$p + q = \frac{- b}{a}, p q = \frac{c}{a}$ ${(p + q)}^{2} - 2 p q = {(p + q)}^{3} - 3 p q (p + q)$ $\Rightarrow \frac{b^{2}}{a^{2}} - \frac{2 c}{a} = \frac{- b^{3}}{a^{3}} + \frac{3 b c}{a^{2}} \Rightarrow {a b}^{2} - 2 a^{2} c = - b^{3} + 3 a b c$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum