Question

hey guys please solve that......​

Answer 1

Solution :-

Given equation :-

ax² + bx + c = 0

Also provided that one root is square of another.

Then let the roots be

$= \alpha \: and \: \alpha^2$

Now as we have :-

Sum of roots :-

$\dfrac{-b}{a} = (\alpha + \alpha^2)$

Product of roots :-

$\dfrac{c}{a} = \alpha^3$

Now we will cube the sum of roots :-

$\rightarrow \left(\dfrac{-b}{a}\right)^3 = (\alpha + \alpha^2)^3$

$\rightarrow \dfrac{-b^3}{a^3} = \alpha^3 + \alpha^6 + 3\alpha^3(\alpha + \alpha^2)$

Now Substitute the sum and product of roots :-

$\rightarrow \dfrac{-b^3}{a^3} = \dfrac{c}{a} + \dfrac{c^2}{a^2} + 3.\dfrac{c}{a} .\dfrac{-b}{a}$

Now multiply both sides by a³

$\rightarrow a^3 \left(\dfrac{-b^3}{a^3} \right)= a^3\left(\dfrac{c}{a} + \dfrac{c^2}{a^2} - 3.\dfrac{bc}{a^2}\right)$

$\rightarrow -b^3 = a^2c + ac^2 - 3abc$

$\rightarrow b^3 + a^2c + ac^2 = 3abc$

Hence Proved !

Answer 2

Answer:

Hey mate your answer is here

Plzzz see above picture..