If a,b,c,d∈R a+b=8 ab+c+d=23 ad+bc=28 cd=12 Find a^2 +b^2 +c^2 +d^2 .


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Question Number 107169 by ZiYangLee last updated on 09/Aug/20

$If a, b, c, d \in R$ $a + b = 8$ $ab + c + d = 23$ $ad + bc = 28$ $cd = 12$ $Find a^{2} + b^{2} + c^{2} + d^{2} .$
Commented by ZiYangLee last updated on 09/Aug/20

$Anyone ?$
	Answered by Her_Majesty last updated on 09/Aug/20

	$(1) a = 8 - b$ $- b^{2} + 8 b + c + d = 23$ $b c - b d + 8 d = 28$ $c d = 12$ $(2) d = 23 + b^{2} - 8 b - c$ $2 c (b - 4) - b^{3} + 16 b^{2} - 87 b + 184 = 28$ $(b^{2} - 8 b - c + 23) c = 12$ $(3) c = \frac{b^{2} - 12 b + 39}{2}$ $b^{4} - 16 b^{3} + 94 b^{2} - 240 b + 225 = 0$ ${(b - 5)}^{2} {(b - 3)}^{2} = 0$ $b_{1} = 3 a_{1} = 5 c_{1} = 6 d_{1} = 2$ $b_{2} = 5 a_{2} = 3 c_{2} = 2 d_{2} = 6$ $a^{2} + b^{2} + c^{2} + d^{2} = 74$
	Answered by 1549442205PVT last updated on 09/Aug/20
	$From the hypothesis we get { ((a+b=8 (1))),((ab+c+d=23(2))),((ad+bc=28(3))),((cd=12(4))) :} Replace (4) into (2)(3)we get { ((ab+((12)/c)+c=23)),((((12a)/c)+bc=28)) :}⇔ { ((abc+c^2 +12=23c)),((12a+bc^2 =28c)) :}(5) Substituting b=12−a into (5)we get { (((8−a)ac+c^2 +12=23c)),((12a+(8−a)c^2 =28c)) :}⇔ { ((8ac−a^2 c+c^2 +12=23c(6))),((12a+8c^2 −ac^2 =28c(7))) :} From (7)we have :a=((8c^2 −28c)/(c^2 −12))⇒a^2 =((64c^4 −448c^3 +784c^2 )/(c^4 −24c^2 +144)) Replace into (7) we obtain: ((8c(8c^2 −28c))/(c^2 −12))−((c(64c^4 −448c^3 +784c^2 ))/(c^4 −24c^2 +144))+c^2 +12=23c ⇔c^6 −12c^4 −144c^2 +1728+64c^5 −224c^4 −768c^3 +2688c^2 −64c^5 +448c^4 −784c^3 =23c^5 −552c^3 +3312c ⇔c^6 −23c^5 +212c^4 −1000c^3 +2544c^2 −3312c+1728 =(c−2)^2 (c−3)(c−4)(c−6)^2 =0 ⇔c∈{2,3,4,6}⇒d∈{6,4,3,2} ⇒a∈{3,4,4,5},b∈{5,4,4,3} ⇒(a,b,c,d)∈{(3,5,2,6),(4,4,3,4),(4,4,4,3),(5,3,6,2)} We find out two different values of S= a^2 +b^2 +c^2 +d^2 being S=4^2 +4^2 +3^2 +4^2 =57 and S=5^2 +3^2 +6^2 +2^2 =74$
	$From the hypothesis we get$ ${\begin{cases} a + b = 8 (1) \\ ab + c + d = 23 (2) \\ ad + bc = 28 (3) \\ cd = 12 (4) \end{cases}$ $Replace (4) into (2) (3) we get$ ${\begin{cases} ab + \frac{12}{c} + c = 23 \\ \frac{12 a}{c} + bc = 28 \end{cases} \Leftrightarrow {\begin{cases} abc + c^{2} + 12 = 23 c \\ 12 a + {bc}^{2} = 28 c \end{cases} (5)$ $Substituting b = 12 - a into (5) we get$ ${\begin{cases} (8 - a) ac + c^{2} + 12 = 23 c \\ 12 a + (8 - a) c^{2} = 28 c \end{cases} \Leftrightarrow {\begin{cases} 8 ac - a^{2} c + c^{2} + 12 = 23 c (6) \\ 12 a + {8 c}^{2} - {ac}^{2} = 28 c (7) \end{cases}$ $From (7) we have : a = \frac{{8 c}^{2} - 28 c}{c^{2} - 12} \Rightarrow a^{2} = \frac{{64 c}^{4} - {448 c}^{3} + {784 c}^{2}}{c^{4} - {24 c}^{2} + 144}$ $Replace into (7) we obtain :$ $\frac{8 c ({8 c}^{2} - 28 c)}{c^{2} - 12} - \frac{c ({64 c}^{4} - {448 c}^{3} + {784 c}^{2})}{c^{4} - {24 c}^{2} + 144} + c^{2} + 12 = 23 c$ $\Leftrightarrow c^{6} - {12 c}^{4} - {144 c}^{2} + 1728 + {64 c}^{5} - {224 c}^{4}$ $- {768 c}^{3} + {2688 c}^{2} - {64 c}^{5} + {448 c}^{4} - {784 c}^{3}$ $= {23 c}^{5} - {552 c}^{3} + 3312 c$ $\Leftrightarrow c^{6} - {23 c}^{5} + {212 c}^{4} - {1000 c}^{3} + {2544 c}^{2} - 3312 c + 1728$ $= {(c - 2)}^{2} (c - 3) (c - 4) {(c - 6)}^{2} = 0$ $\Leftrightarrow c \in {2, 3, 4, 6} \Rightarrow d \in {6, 4, 3, 2}$ $\Rightarrow a \in {3, 4, 4, 5}, b \in {5, 4, 4, 3}$ $\Rightarrow (a, b, c, d) \in {(3, 5, 2, 6), (4, 4, 3, 4), (4, 4, 4, 3), (5, 3, 6, 2)}$ $We find out two different values of$ $S = a^{2} + b^{2} + c^{2} + d^{2} being$ $S = 4^{2} + 4^{2} + 3^{2} + 4^{2} = 57$ $and S = 5^{2} + 3^{2} + 6^{2} + 2^{2} = 74$
	Commented by Her_Majesty last updated on 09/Aug/20

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