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Question Number 202250 by MATHEMATICSAM last updated on 23/Dec/23

(a^{2} - b c) x^{2} + 2 (b^{2} - c a) x + (c^{2} - a b) = 0

has two equal roots . Show that either

b = 0 or \frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} = 3 .

Answered by esmaeil last updated on 23/Dec/23

\underset{1}{r} - r_{2} \overset{r_{1} = \underset{2}{r}}{=} \frac{\sqrt{δ^{^{'}}}}{∣ a ∣} = 0 \to

\frac{\sqrt{{(b^{2} - c a)}^{2} - (c^{2} - a b) (a^{2} - b c)}}{a^{2} - b c} = 0 \to

b^{4} + c^{2} a^{2} - 2 b^{2} a c - c^{2} a^{2} + {b c}^{3} + {b a}^{3} - {a b}^{2} c

= 0

\to b^{4} - 3 b^{2} a c + {b c}^{3} + {b a}^{3} ==\to

b (b^{3} - 3 b a c + c^{3} + a^{3}) = 0 \to

b = 0 o r \frac{b^{3}}{a b c} - \frac{3 a b c}{a b c} + \frac{c^{3}}{a b c} + \frac{{b a}^{3}}{a b c} = 0 \to

\frac{b^{2}}{a c} + \frac{c^{2}}{a b} + \frac{a^{2}}{b c} = 3

Answered by Rasheed.Sindhi last updated on 23/Dec/23

α & α a r e t h e r o o t s

α + α = 2 α = - \frac{2 (b^{2} - c a)}{(a^{2} - b c)} \Rightarrow α = \frac{b^{2} - c a}{a^{2} - b c}

α^{2} = {(\frac{b^{2} - c a}{a^{2} - b c})}^{2} = \frac{c^{2} - a b}{a^{2} - b c}

\frac{{(b^{2} - c a)}^{2}}{a^{2} - b c} = c^{2} - a b

{(b^{2} - c a)}^{2} = (a^{2} - b c) (c^{2} - a b)

b^{4} - 2 {a b}^{2} c + c^{2} a^{2} = c^{2} a^{2} - a^{3} b - {b c}^{3} + {a b}^{2} c

b^{4} - 3 {a b}^{2} c = - a^{3} b - {b c}^{3}

b^{4} - 3 {a b}^{2} c + a^{3} b + {b c}^{3} = 0

b (b^{3} - 3 a b c + a^{3} + c^{3}) = 0

b = 0 ✓ ∣ a^{3} + b^{3} + c^{3} = 3 a b c

\frac{a^{3} + b^{3} + c^{3}}{a b c} = 3

\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} = 3 ✓