Question

if the sum of the roots of the quadratic equation ax²+bx+c=0 is equal to the sum of the cubes of their reciprocals then prove that ab²=3a²c+c³​

Answer 1

Answer:

let roots be α and β

A= a B= b and C=c

α+β= -B/A

⇒α+β= - b/a   --(1)

αβ=C/A

⇒αβ=c/a     --(2)

According to question

α+β= 1/α³+1/β³= -b/a

(α³+β³)/(αβ)³=-b/a

(α +β)( α²+β² - αβ)= -b(αβ)³/a

(-b/a)((α+β)²-2αβ-αβ)= -b(αβ)³/a

(-b/a)²-3c/a= (c/a)³  -(from (1) and (2))

b²/a²-3c/a= c³/a³

b²a- 3ca³= c³

ab²= c³ + 3ca³

hence proved

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Answer 2

Answer:

Pls find in attachment

Step-by-step explanation:

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