If-and-are-the-roots-of-ax-3-bx-c-0-then-frame-an-equation-whose-roots-are-2-2-2-

Question Number 202436 by MATHEMATICSAM last updated on 26/Dec/23

If α, β and γ are the roots of

{a x}^{3} + b x + c = 0 then frame an equation

whose roots are α^{2}, β^{2}, γ^{2} .

Answered by aleks041103 last updated on 26/Dec/23

{a x}^{3} + b x + c = 0, x = α, β, γ

\Rightarrow - \frac{c}{a} = α β γ

\frac{b}{a} = α β + β γ + α γ

0 = α + β + γ

{A x}^{3} + {B x}^{2} + C x + D = 0, x = α^{2}, β^{2}, γ^{2}

- \frac{D}{A} = α^{2} β^{2} γ^{2} = {(α β γ)}^{2} = \frac{c^{2}}{a^{2}}

\frac{C}{A} = α^{2} β^{2} + β^{2} γ^{2} + α^{2} γ^{2} = {(α β + β γ + α γ)}^{2} - 2 α β γ (α + β + γ) =

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

- \frac{B}{A} = α^{2} + β^{2} + γ^{2} = {(α + β + γ)}^{2} - 2 (α β + β γ + α γ) =

= 0^{2} - 2 \frac{b}{a} = - \frac{2 b}{a}

x^{3} + \frac{2 b}{a} x^{2} + \frac{b^{2}}{a^{2}} x - \frac{c^{2}}{a^{2}} = 0

\Rightarrow a^{2} x^{3} + 2 {a b x}^{2} + b^{2} x - c^{2} = 0

Answered by Frix last updated on 27/Dec/23

x^{3} + {p x}^{2} + q x + r = 0

(\sqrt{x^{3}} + p x + q \sqrt{x} + r) (\sqrt{x^{3}} - p x + q \sqrt{x} - r) = 0

x^{3} - (p^{2} - 2 q) x^{2} - (2 p r - q^{2}) x - r^{2} = 0

p = 0 \land q = \frac{b}{a} \land r = \frac{c}{a}

x^{3} + \frac{2 {b x}^{2}}{a} + \frac{b^{2} x}{a^{3}} - \frac{c^{2}}{a^{2}} = 0

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

Facebook Tweet Pin