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-

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

{\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

y o u a r e r i g h t \dots b u t I {d o n}^{'} t f u l l y u n d e r s t a n d

w h e r e / w h y I^{'} m l o s i n g 2 s o l u t i o n s