If-b-3-a-2-c-ac-2-3abc-then-prove-that-one-root-of-ax-2-bx-c-0-is-the-square-of-the-other-one-

Question Number 215837 by MATHEMATICSAM last updated on 19/Jan/25

If b^{3} + a^{2} c + {a c}^{2} = 3 a b c then prove that

one root of {a x}^{2} + b x + c = 0 is the square

of the other one .

Answered by MrGaster last updated on 19/Jan/25

given b^{3} + a^{2} c + {a c}^{2} = 3 a b c

β b e the roots of {a x}^{2} + b x + c = 0

by {Vieta}^{'} s formulas : Σ α = - \frac{b}{a} \land Π α = \frac{c}{a}

if one root is the of the other, then \exists γ : α = γ^{2} \land β = γ \lor α = γ \land β = γ^{2}

without loss of generality, assume α = γ^{2} \land β = γ

∴ Σ α = γ^{2} + γ = - \frac{b}{a}

Π α = γ^{3} = \frac{c}{a}

from beth sides by a^{3} : {(\frac{b}{a})}^{3} + (\frac{c}{a}) + {(\frac{c}{a})}^{2} = 3 (\frac{b}{a}) (\frac{c}{a})

substitute Σ α and Π α : {(- Σ α)}^{3} + Π α + {(Π α)}^{2} = 3 (- Σ α) (Π α)

{(- (γ^{2} + γ))}^{3} + γ^{3} + {(γ^{3})}^{2} = 3 (- (γ^{2} + γ)) (γ^{3})

- 3 γ^{4} = - 3 γ^{4}

b x + c = 0 is indeed the of the other .

ps : “^{″} denotes squaring .

Commented by A5T last updated on 19/Jan/25

From the wording of the question, what is

required to prove is :

b^{3} + a^{2} c + {ac}^{2} = 3 abc \Rightarrow x_{1} = γ \land x_{2} = γ^{2}

You showed the converse here :

x_{1} = γ \land x_{2} = γ^{2} \Rightarrow b^{3} + a^{2} c + {ac}^{2} = 3 abc

Both are not quite the same .

A \Rightarrow B ≢ B \Rightarrow A .

Answered by A5T last updated on 19/Jan/25

Let the roots of {ax}^{2} + bx + c = 0 be x_{1} and x_{2}

x_{1} + x_{2} = \frac{- b}{a}; x_{1} x_{2} = \frac{c}{a}

b^{3} + a^{2} c + {ac}^{2} = 3 abc \overset{/ a^{3}}{\Rightarrow} {(\frac{b}{a})}^{3} + (\frac{c}{a}) + {(\frac{c}{a})}^{2} = 3 (\frac{b}{a}) (\frac{c}{a})

\Rightarrow - {(x_{1} + x_{2})}^{3} + x_{1} x_{2} + x_{1}^{2} x_{2}^{2} = - 3 (x_{1} + x_{2}) (x_{1} x_{2})

\Rightarrow - (x_{1}^{3} + {3 x}_{1}^{2} x_{2} + {3 x}_{1} x_{2}^{2} + x_{2}^{3}) + x_{1} x_{2} + x_{1}^{2} x_{2}^{2} = - {3 x}_{1}^{2} x_{2} - {3 x}_{1} x_{2}^{2}

\Rightarrow - x_{1}^{3} - x_{2}^{3} + x_{1} x_{2} + x_{1}^{2} x_{2}^{2} = - x_{1}^{2} (x_{1} - x_{2}^{2}) + x_{2} (x_{1} - x_{2}^{2}) = 0

\Rightarrow (x_{1}^{2} - x_{2}) (x_{1} - x_{2}^{2}) = 0

\Rightarrow x_{1}^{2} = x_{2} or x_{2}^{2} = x_{1}

Commented by ArshadS last updated on 19/Jan/25

5 t h l i n e :

\Rightarrow - (x_{1}^{3} + {3 x}_{1}^{2} x_{2} + {3 x}_{1} x_{2}^{2} + x_{2}^{3}) + x_{1} x_{2} + x_{1}^{2} x_{2}^{2} = - {3 x}_{1}^{2} x_{2} - {3 x}_{1} x_{2}^{2}

Commented by A5T last updated on 19/Jan/25

It was a typo, thanks .

Facebook Tweet Pin