in question 34992 this had to be solved: t^4 +8t^3 +2t^2 −8t+1=0 here another concept worked (I tried some) t_1 =a+b+c+d t_2 =a+b−c−d t_3 =a−b+c−d t_3 =a−b−c+d (t−t_1 )(t−t_2 )(t−t_3 )(t−t_4 )=0 leads to t^4 −4at^3 +2(3a^2 −b^2 −c^2 −d^2 )t^2 + +4(a(−a^2 +b^2 +c^2 +d^2 )−2bcd)t+ +a^4 +b^4 +c^4 +d^4 −2(a^2 (b^2 +c^2 +d^2 )+b^2 (c^2 +d^2 )+c^2 d^2 ) 1. −4a=8 2. 2(3a^2 −b^2 −c^2 −d^2 )=2 3. 4(a(−a^2 +b^2 +c^2 +d^2 )−2bcd)=−8 4. a^4 +b^4 +c^4 +d^4 −2(a^2 (b^2 +c^2 +d^2 )+b^2 (c^2 +d^2 )+c^2 d^2 )=1 1. a=−2 2. b^2 +c^2 +d^2 =11 3. b^2 +c^2 +d^2 +bcd=5 4. ... 3. −2. bcd=−6 now I tried by chance if there are easy real solutions b^2 =p, c^2 =q, d^2 =r, 0≤p≤q≤r p+q+r=11 this leads to p=2, q=3, p=6 ∣b∣=(√2), ∣c∣=(√3), ∣d∣=(√6) with one or three out of a, b, c negative because of bcd=−6 because of the nature of t_1 , t_2 , t_3 , t_4 these lead to the same solutions I took b=−(√2), c=−(√3), d=−(√6) t_1 =−2−(√2)−(√3)−(√6) t_2 =−2−(√2)+(√3)+(√6) t_3 =−2+(√2)−(√3)+(√6) t


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Question Number 36219 by MJS last updated on 30/May/18

$in question 34992 this had to be solved :$ $t^{4} + 8 t^{3} + 2 t^{2} - 8 t + 1 = 0$ $here another concept worked (I tried some)$ $t_{1} = a + b + c + d$ $t_{2} = a + b - c - d$ $t_{3} = a - b + c - d$ $t_{3} = a - b - c + d$ $(t - t_{1}) (t - t_{2}) (t - t_{3}) (t - t_{4}) = 0$ $leads to$ $t^{4} - 4 {a t}^{3} + 2 (3 a^{2} - b^{2} - c^{2} - d^{2}) t^{2} +$ $+ 4 (a (- a^{2} + b^{2} + c^{2} + d^{2}) - 2 b c d) t +$ $+ a^{4} + b^{4} + c^{4} + d^{4} - 2 (a^{2} (b^{2} + c^{2} + d^{2}) + b^{2} (c^{2} + d^{2}) + c^{2} d^{2})$ $1 . - 4 a = 8$ $2 . 2 (3 a^{2} - b^{2} - c^{2} - d^{2}) = 2$ $3 . 4 (a (- a^{2} + b^{2} + c^{2} + d^{2}) - 2 b c d) = - 8$ $4 . a^{4} + b^{4} + c^{4} + d^{4} - 2 (a^{2} (b^{2} + c^{2} + d^{2}) + b^{2} (c^{2} + d^{2}) + c^{2} d^{2}) = 1$ $1 . a = - 2$ $2 . b^{2} + c^{2} + d^{2} = 11$ $3 . b^{2} + c^{2} + d^{2} + b c d = 5$ $4 . . . .$ $3 . - 2 . b c d = - 6$ $now I tried by chance if there are easy real solutions$ $b^{2} = p, c^{2} = q, d^{2} = r, 0 ⩽ p ⩽ q ⩽ r$ $p + q + r = 11$ $this leads to$ $p = 2, q = 3, p = 6$ $∣ b ∣= \sqrt{2}, ∣ c ∣= \sqrt{3}, ∣ d ∣= \sqrt{6} with one or three out$ $of a, b, c negative because of b c d = - 6$ $because of the nature of t_{1}, t_{2}, t_{3}, t_{4} these$ $lead to the same solutions$ $I took b = - \sqrt{2}, c = - \sqrt{3}, d = - \sqrt{6}$ $t_{1} = - 2 - \sqrt{2} - \sqrt{3} - \sqrt{6}$ $t_{2} = - 2 - \sqrt{2} + \sqrt{3} + \sqrt{6}$ $t_{3} = - 2 + \sqrt{2} - \sqrt{3} + \sqrt{6}$ $t_{4} = - 2 + \sqrt{2} + \sqrt{3} - \sqrt{6}$
Commented by Rasheed.Sindhi last updated on 30/May/18
Creative work , I think! �� ✌
Commented by tanmay.chaudhury50@gmail.com last updated on 31/May/18

$t^{4} + 8 t^{3} + 2 t^{2} - 8 t + 1 = 0$ $t^{2} + 8 t + 2 - \frac{8}{t} + \frac{1}{t^{2}} = 0$ $t^{2} + \frac{1}{t^{2}} + 8 (t - \frac{1}{t}) + 2 = 0$ ${(t - \frac{1}{t})}^{2} + 2 + 8 (t - \frac{1}{t}) + 2 = 0$ $k = t - \frac{1}{t}$ $k^{2} + 8 k + 4 = 0$ $k = \frac{- 8 \pm \sqrt{64 - 16}}{2}$ $= \frac{- 8 \pm 4 \sqrt{3}}{2}$ $= - 4 \pm 2 \sqrt{3}$ $k = - 4 + 2 \sqrt{3} a n d - 4 - 2 \sqrt{3}$ $n o w$ $t - \frac{1}{t} = k$ $t^{2} - 1 = t k$ $t^{2} - t k - 1 = 0$ $t = \frac{k \pm \sqrt{k^{2} + 4}}{2}$ $= \frac{k + \sqrt{k^{2} + 4}}{2} a n d \frac{k - \sqrt{k^{2} + 4}}{2}$ $n o w p u t v a l u e o f k = - 4 + 2 \sqrt{3}$ $t = \frac{- 4 + 2 \sqrt{3} \pm \sqrt{16 - 16 \sqrt{3} + 12 + 4}}{2}$ $t = \frac{- 4 + 2 \sqrt{3} \pm \sqrt{32 - 16 \sqrt{3}}}{2}$ $t = - 2 + \sqrt{3} \pm \sqrt{8 - 4 \sqrt{3}}$ $t = - 2 + \sqrt{3} + \sqrt{8 - 4 \sqrt{3}} a n d - 2 + \sqrt{3} - \sqrt{8 - 4 \sqrt{3}}$ $n o w p u t k = - 4 - 2 \sqrt{3} y o u w i l l g e t t w o v a l u e o f$ $t s o t o t a l 2 + 2 = 4 r o o t s o f t f o u n d$
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