x^4 +10x^3 +6x−1=0 for those who want an exact solution... (x−a−bi)(x−a+bi)(x−c−d)(x−c+d)=0 x^4 −2(a+c)x^3 +(a^2 +4ac+b^2 +c^2 −d^2 )x^2 − −2(a(c^2 −d^2 )+c(a^2 +b^2 ))x+(a^2 +b^2 )(c^2 −d^2 )=0 1. −2(a+c)=10 2. a^2 +4ac+b^2 +c^2 −d^2 =0 3. −2(a(c^2 −d^2 )+c(a^2 +b^2 ))=6 4. (a^2 +b^2 )(c^2 −d^2 )=−1 1. a=−c−5e 2. b=(√(−a^2 +4ac−c^2 +d^2 ))=(√(2c^2 +10c+d^2 −25)) 3. d=(√(ac+c^2 +((b^2 c+3)/a)))=(√(−((2c^3 +15c^2 +3)/(2c+5)))) 2. b=(√((2c^3 +15c^2 −128)/(2c+5))) 4. 4c^4 +40c^3 +75c^2 −125c+((689)/4)−((17161)/(4(2c+5)^2 ))=0 64(c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1)=0 c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1=0 c=u−((15)/6) u^6 −((75)/4)u^4 +((1939)/(16))u^2 −((17161)/(64))=0 u=(√v) v^3 −((75)/4)v^2 +((1939)/(16))v−((17161)/(64))=0 v=w+((75)/(12)) w^3 +4w+1=0 w=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) +((−(q/2)−(√((p^3 /(27))+(q^2 /4)))))^(1/3) = =((−(1/2)+((√(849))/(18))))^(1/3) +((−(1/2)−((√(849))/(18))))^(1/3) and now we have to go all the way back... not very friendly... as I mentioned before, it′s almost (or maybe absolutely) impossible to find a nicer exact form of w, so let me use the approximation w≈−.246266 v=((25)/4)+w≈6.00373 u=((√(25+4w))/2)≈2.45025 c=−(5/2)+((√(25+4w))/2)≈−.0497482 d=(√(((25)/2)−w−((131)/(2(√(25+4w))))))≈.787215i a=−(5/2)−((√(25+4w))/2)≈−4.95025 b=(√(−((25)/2)+w+((131)/(2(√(25+4w))))))≈5.11001i x_1 =a+bi≈−10.0603 x_2 =a−bi≈.159762 x_3 =c+d≈−.0497482+.787215i x


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Question Number 36207 by MJS last updated on 30/May/18

$x^{4} + 10 x^{3} + 6 x - 1 = 0$ $for those who want an exact solution . . .$ $(x - a - b i) (x - a + b i) (x - c - d) (x - c + d) = 0$ $x^{4} - 2 (a + c) x^{3} + (a^{2} + 4 a c + b^{2} + c^{2} - d^{2}) x^{2} -$ $- 2 (a (c^{2} - d^{2}) + c (a^{2} + b^{2})) x + (a^{2} + b^{2}) (c^{2} - d^{2}) = 0$ $1 . - 2 (a + c) = 10$ $2 . a^{2} + 4 a c + b^{2} + c^{2} - d^{2} = 0$ $3 . - 2 (a (c^{2} - d^{2}) + c (a^{2} + b^{2})) = 6$ $4 . (a^{2} + b^{2}) (c^{2} - d^{2}) = - 1$ $1 . a = - c - 5 e$ $2 . b = \sqrt{- a^{2} + 4 a c - c^{2} + d^{2}} = \sqrt{2 c^{2} + 10 c + d^{2} - 25}$ $3 . d = \sqrt{a c + c^{2} + \frac{b^{2} c + 3}{a}} = \sqrt{- \frac{2 c^{3} + 15 c^{2} + 3}{2 c + 5}}$ $2 . b = \sqrt{\frac{2 c^{3} + 15 c^{2} - 128}{2 c + 5}}$ $4 .$ $4 c^{4} + 40 c^{3} + 75 c^{2} - 125 c + \frac{689}{4} - \frac{17161}{4 {(2 c + 5)}^{2}} = 0$ $64 (c^{6} + 15 c^{5} + 75 c^{4} + 125 c^{3} + 4 c^{2} + 20 c + 1) = 0$ $c^{6} + 15 c^{5} + 75 c^{4} + 125 c^{3} + 4 c^{2} + 20 c + 1 = 0$ $c = u - \frac{15}{6}$ $u^{6} - \frac{75}{4} u^{4} + \frac{1939}{16} u^{2} - \frac{17161}{64} = 0$ $u = \sqrt{v}$ $v^{3} - \frac{75}{4} v^{2} + \frac{1939}{16} v - \frac{17161}{64} = 0$ $v = w + \frac{75}{12}$ $w^{3} + 4 w + 1 = 0$ $w = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} =$ $= \sqrt[3]{- \frac{1}{2} + \frac{\sqrt{849}}{18}} + \sqrt[3]{- \frac{1}{2} - \frac{\sqrt{849}}{18}}$ $and now we have to go all the way back . . .$ $not very friendly . . .$ $as I mentioned before, {it}^{'} s almost (or maybe$ $absolutely) impossible to find a nicer exact$ $form of w, so let me use the approximation$ $w \approx - . 246266$ $v = \frac{25}{4} + w \approx 6.00373$ $u = \frac{\sqrt{25 + 4 w}}{2} \approx 2.45025$ $c = - \frac{5}{2} + \frac{\sqrt{25 + 4 w}}{2} \approx - . 0497482$ $d = \sqrt{\frac{25}{2} - w - \frac{131}{2 \sqrt{25 + 4 w}}} \approx . 787215 i$ $a = - \frac{5}{2} - \frac{\sqrt{25 + 4 w}}{2} \approx - 4.95025$ $b = \sqrt{- \frac{25}{2} + w + \frac{131}{2 \sqrt{25 + 4 w}}} \approx 5 . 11001 i$ $x_{1} = a + b i \approx - 10.0603$ $x_{2} = a - b i \approx . 159762$ $x_{3} = c + d \approx - . 0497482 + . 787215 i$ $x_{4} = c - d \approx - . 0497482 - . 787215 i$
Commented by abdo mathsup 649 cc last updated on 30/May/18

$t h a n k y o u s i r M j s$
Commented by Rasheed.Sindhi last updated on 30/May/18

$^{V} N i c e S i r!$
Commented by behi83417@gmail.com last updated on 30/May/18

$t h a n k s i n a d v a n c e d e a r m r : M J S .$
Commented by MJS last updated on 30/May/18

$you are welcome$
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