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Question Number 36132 by Rasheed.Sindhi last updated on 29/May/18

a + b = 10 \dots \dots \dots (i)

a b + c = 0 \dots \dots \dots . (ii)

a c + d = 6 \dots \dots \dots . (iii)

a d = - 1 \dots \dots \dots . . (iv)

(a, b, c, d) = ?

N o t e : T h i s p r o b l e m i s r e l a t e d t o s o l v e

t h e e q u a t i o n (t^{4} + 10 t + 6 t - 1 = 0) o f

You can't use 'macro parameter character #' in math mode

Commented by Rasheed.Sindhi last updated on 29/May/18

(i) \Rightarrow a = 10 - b \Rightarrow

(ii) : (10 - b) b + c = 0 \Rightarrow c = (b - 10) b

(iii) : (10 - b) c + d = 6 \Rightarrow (10 - b) {(b - 10) b} = 6

(iv) : (10 - b) d = - 1

(ii) : c = (b - 10) b \Rightarrow

(iii) : (10 - b) (b - 10) b - \frac{1}{10 - b} = 6

- (b - 10) (b - 10) b + \frac{1}{b - 10} = 6

- b {(b - 10)}^{3} + 1 = 6 (b - 10)

W rongly derived cubic equation . Sorry!

b^{3} - 20 b^{2} + 100 b + 6 = 0

A unybody who know how to solve cubic equation . .

Commented by prakash jain last updated on 29/May/18

There is a procedure to solve cubic

equation and also cubic formula .

It is easier to follow step then remembering

the formula .

Commented by mondodotto@gmail.com last updated on 29/May/18

use a calculator

Commented by prakash jain last updated on 29/May/18

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Commented by Rasheed.Sindhi last updated on 29/May/18

T h \infty n k S S i r! I a l w a y s h a v e l e a r n t

f r o m y o u .

Commented by MJS last updated on 29/May/18

the problem with {Cardano}^{'} s method is,

you {won}^{'} t get real solutions of usable form

(x - 3) (x + 4) (x - 5) = 0

x^{3} - 4 x^{2} - 17 x + 60 = 0

x = z + \frac{4}{3}

z^{3} - \frac{67}{3} z + \frac{880}{27} = 0

p = - \frac{67}{3}; q = \frac{880}{27}

z_{1} = u + v

z_{2} = (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) v

z_{3} = (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) v

u = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} = \sqrt[3]{- \frac{440}{27} + 7 \sqrt{3} i}

v = \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} = \sqrt[3]{- \frac{440}{27} + 7 \sqrt{3} i}

u + v = \sqrt[3]{- \frac{440}{27} + 7 \sqrt{3} i} + \sqrt[3]{- \frac{440}{27} + 7 \sqrt{3} i} = \frac{11}{3}

but you cannot easily show this is true

and in a case like

(x - a - b i) (x - a + b i) (x - c) = 0

{it}^{'} s impossible

Commented by Rasheed.Sindhi last updated on 30/May/18

T h a n k S a l o t S i r!