Find-the-condition-that-one-root-of-ax-2-bx-c-0-a-0-is-square-of-the-other-

Question Number 132697 by liberty last updated on 15/Feb/21

Find the condition that one

root of {a x}^{2} + b x + c = 0, a \neq 0

is square of the other .

Commented by liberty last updated on 16/Feb/21

okay

Commented by mr W last updated on 16/Feb/21

α + α^{2} = - \frac{b}{a}

α \times α^{2} = \frac{c}{a}

\Rightarrow α = \sqrt[3]{\frac{c}{a}}

\Rightarrow α (1 + α) = \sqrt[3]{\frac{c}{a}} (1 + \sqrt[3]{\frac{c}{a}}) = - \frac{b}{a}

\Rightarrow \frac{b}{a} + \sqrt[3]{\frac{c}{a}} (1 + \sqrt[3]{\frac{c}{a}}) = 0

Answered by EDWIN88 last updated on 16/Feb/21

Suppose υ is a root and that the other one

is υ^{2} . By {Vietha}^{'} s rule {\begin{cases} υ + υ^{2} = - \frac{b}{a} \\ υ^{3} = \frac{c}{a} \end{cases}

W e have υ = 0 if only if b = c = 0

let we assume that υ \neq 0

Therefore \frac{1 + υ}{υ^{2}} = - \frac{b}{c} \Rightarrow b υ^{2} + c υ + c = 0

on the other hand we find a υ^{2} + b υ + c = 0

substract the two relations we get

(a - b) υ^{2} + (b - c) υ = 0

(a - b) υ + (b - c) = 0 \to υ = \frac{c - b}{a - b}

b u t w e c a n t h e t w o r e l a t i o n s m u l t i p l y i n g

t h e t o p o n e b y a a n d t h e b o t t o m b y b

{\begin{cases} a b υ^{2} + a c υ + a c = 0 \\ a b υ^{2} + b^{2} υ + b c = 0 \end{cases}

s u b t r a c t (a c - b^{2}) υ + (a c - b c) = 0 \to υ = \frac{b c - a c}{a c - b^{2}}

t h a t e x p a n d s t o a^{2} c + {a c}^{2} - 3 a b c + b^{3} = 0

Commented by liberty last updated on 16/Feb/21

okay

Commented by EDWIN88 last updated on 16/Feb/21

let for case {\begin{cases} a = 1; b = - 2 \\ c + c^{2} + 6 c - 8 = 0 \Rightarrow c^{2} + 7 c - 8 = 0 \end{cases}

(c + 8) (c - 1) = 0 {\begin{cases} c = - 8 \\ c = 1 \end{cases}

If (a, b, c) = (1, - 2, - 8) \Rightarrow x^{2} - 2 x - 8 = 0

(x - 4) (x + 2) = 0 \Rightarrow x_{1} = 4 \land x_{2} = - 2

and {(- 2)}^{2} = 4

Answered by MJS_new last updated on 16/Feb/21

a (x - α^{2}) (x - α) = 0

{a x}^{2} - a α (α + 1) x + a α^{3} = 0

b = - a α (α + 1)

c = a α^{3}

not sure what else you need \dots

Commented by liberty last updated on 16/Feb/21

relation for a, b and c sir

Commented by MJS_new last updated on 16/Feb/21

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

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

(x - α^{2}) (x - α) = x^{2} + p x + q

\Leftrightarrow

p = - α (α + 1) \land q = α^{3}

\Leftrightarrow

p + q^{2 / 3} + q^{1 / 3} = 0

\Leftrightarrow

p = - (1 + q^{1 / 3}) q^{1 / 3} \Leftrightarrow q = \frac{3 p - 1 \pm (p - 1) \sqrt{1 - 4 p}}{2}

or

p^{3} - 3 p q + q^{2} + q = 0

with p = \frac{b}{a} \land q = \frac{c}{a} we get

a = \frac{3 b - c}{2} \pm \frac{b - c}{2} \sqrt{1 - \frac{4 b}{c}}

\Leftrightarrow

b = a^{1 / 3} (a^{1 / 3} + c^{1 / 3}) c^{1 / 3}

\Leftrightarrow

c = \frac{3 b - a}{2} \pm \frac{b - a}{2} \sqrt{1 - \frac{4 b}{a}}

or

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

Answered by Rasheed.Sindhi last updated on 16/Feb/21

{\begin{cases} α + α^{2} = - \frac{b}{a} \\ α^{3} = \frac{c}{a} \end{cases}

{(α + α^{2})}^{3} = {(- \frac{b}{a})}^{3}

α^{3} + {(α^{3})}^{2} + 3 α^{3} (α + α^{2}) = - \frac{b^{3}}{a^{3}}

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

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

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

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