If one of the roots of the equation \( \Large x^{2}-bx+c=0 \) is the square of the other, then which of the following option is correct?

A) \( \Large b^{2} = 3bc+c^{2}+c \)

B) \( \Large c^{3} = 3bc+b^{2}+b \)

C) \( \Large 3bc = c^{3}+b^{2}+b \)

D) \( \Large 3bc = c^{3}+b^{3}+b^{2} \)

Correct answer:
A) \( \Large b^{2} = 3bc+c^{2}+c \)

Description for Correct answer:

Given that, one root of the equation \( \Large x^{2} - bx + c = 0\) is square of other root of this equation i.e., roots (\( \Large \left( \alpha , \alpha ^{2}\right) \)).

Sum of roots = \( \Large \alpha + \alpha ^{2} = - \frac{-b}{1} \)

= \( \Large \alpha \left( \alpha + 1\right) = b \) ...(i)

and product of roots = \( \Large \alpha . \alpha ^{2} = \frac{c}{1} \)

=> \( \Large \alpha ^{3} = c => \alpha = c^{1/3} \) ...(ii)

From Eqs. (i) and (ii),

\( \Large c^{1/3} \left(c^{1/3} + 1\right) = b \) ...(iii)

On ubing both sides, we get

\( \Large c \left(c^{1/3} + 1\right)^{3} = b^{3} \)

=>\( \Large c \{c + 1 + 3c^{1/3} \left(c^{1/3} + 1 \right) \} = b^{3} \)

=>\( \Large c [c + 1 + 3b] = b^{3} \) [From Eq. (iii)]

\( \Large b^{3} = 3bc + c^{2} + c \)



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