Menu Close

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-

Tinku Tara 0 Comments
Pin




Question Number 107169 by ZiYangLee last updated on 09/Aug/20
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 .
$$\mathrm{If}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in\mathbb{R} \\ $$$$\mathrm{a}+\mathrm{b}=\mathrm{8} \\ $$$$\mathrm{ab}+\mathrm{c}+\mathrm{d}=\mathrm{23} \\ $$$$\mathrm{ad}+\mathrm{bc}=\mathrm{28} \\ $$$$\mathrm{cd}=\mathrm{12} \\ $$$$\mathrm{Find}\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} . \\ $$$$ \\ $$
Commented by ZiYangLee last updated on 09/Aug/20
Anyone?
$$\mathrm{Anyone}? \\ $$
Answered by Her_Majesty last updated on 09/Aug/20
(1) a=8−b −b^2 +8b+c+d=23 bc−bd+8d=28 cd=12 (2) d=23+b^2 −8b−c 2c(b−4)−b^3 +16b^2 −87b+184=28 (b^2 −8b−c+23)c=12 (3) c=((b^2 −12b+39)/2) b^4 −16b^3 +94b^2 −240b+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
$$\left(\mathrm{1}\right)\:{a}=\mathrm{8}−{b} \\ $$$$−{b}^{\mathrm{2}} +\mathrm{8}{b}+{c}+{d}=\mathrm{23} \\ $$$${bc}−{bd}+\mathrm{8}{d}=\mathrm{28} \\ $$$${cd}=\mathrm{12} \\ $$$$\left(\mathrm{2}\right)\:{d}=\mathrm{23}+{b}^{\mathrm{2}} −\mathrm{8}{b}−{c} \\ $$$$\mathrm{2}{c}\left({b}−\mathrm{4}\right)−{b}^{\mathrm{3}} +\mathrm{16}{b}^{\mathrm{2}} −\mathrm{87}{b}+\mathrm{184}=\mathrm{28} \\ $$$$\left({b}^{\mathrm{2}} −\mathrm{8}{b}−{c}+\mathrm{23}\right){c}=\mathrm{12} \\ $$$$\left(\mathrm{3}\right)\:{c}=\frac{{b}^{\mathrm{2}} −\mathrm{12}{b}+\mathrm{39}}{\mathrm{2}} \\ $$$${b}^{\mathrm{4}} −\mathrm{16}{b}^{\mathrm{3}} +\mathrm{94}{b}^{\mathrm{2}} −\mathrm{240}{b}+\mathrm{225}=\mathrm{0} \\ $$$$\left({b}−\mathrm{5}\right)^{\mathrm{2}} \left({b}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${b}_{\mathrm{1}} =\mathrm{3}\:{a}_{\mathrm{1}} =\mathrm{5}\:{c}_{\mathrm{1}} =\mathrm{6}\:{d}_{\mathrm{1}} =\mathrm{2} \\ $$$${b}_{\mathrm{2}} =\mathrm{5}\:{a}_{\mathrm{2}} =\mathrm{3}\:{c}_{\mathrm{2}} =\mathrm{2}\:{d}_{\mathrm{2}} =\mathrm{6} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{74} \\ $$
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
$$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\begin{cases}{\mathrm{a}+\mathrm{b}=\mathrm{8}\:\left(\mathrm{1}\right)}\\{\mathrm{ab}+\mathrm{c}+\mathrm{d}=\mathrm{23}\left(\mathrm{2}\right)}\\{\mathrm{ad}+\mathrm{bc}=\mathrm{28}\left(\mathrm{3}\right)}\\{\mathrm{cd}=\mathrm{12}\left(\mathrm{4}\right)}\end{cases} \\ $$$$\mathrm{Replace}\:\left(\mathrm{4}\right)\:\mathrm{into}\:\left(\mathrm{2}\right)\left(\mathrm{3}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\begin{cases}{\mathrm{ab}+\frac{\mathrm{12}}{\mathrm{c}}+\mathrm{c}=\mathrm{23}}\\{\frac{\mathrm{12a}}{\mathrm{c}}+\mathrm{bc}=\mathrm{28}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{abc}+\mathrm{c}^{\mathrm{2}} +\mathrm{12}=\mathrm{23c}}\\{\mathrm{12a}+\mathrm{bc}^{\mathrm{2}} =\mathrm{28c}}\end{cases}\left(\mathrm{5}\right) \\ $$$$\mathrm{Substituting}\:\mathrm{b}=\mathrm{12}−\mathrm{a}\:\mathrm{into}\:\left(\mathrm{5}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\begin{cases}{\left(\mathrm{8}−\mathrm{a}\right)\mathrm{ac}+\mathrm{c}^{\mathrm{2}} +\mathrm{12}=\mathrm{23c}}\\{\mathrm{12a}+\left(\mathrm{8}−\mathrm{a}\right)\mathrm{c}^{\mathrm{2}} =\mathrm{28c}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{8ac}−\mathrm{a}^{\mathrm{2}} \mathrm{c}+\mathrm{c}^{\mathrm{2}} +\mathrm{12}=\mathrm{23c}\left(\mathrm{6}\right)}\\{\mathrm{12a}+\mathrm{8c}^{\mathrm{2}} −\mathrm{ac}^{\mathrm{2}} =\mathrm{28c}\left(\mathrm{7}\right)}\end{cases} \\ $$$$\mathrm{From}\:\left(\mathrm{7}\right)\mathrm{we}\:\mathrm{have}\::\mathrm{a}=\frac{\mathrm{8c}^{\mathrm{2}} −\mathrm{28c}}{\mathrm{c}^{\mathrm{2}} −\mathrm{12}}\Rightarrow\mathrm{a}^{\mathrm{2}} =\frac{\mathrm{64c}^{\mathrm{4}} −\mathrm{448c}^{\mathrm{3}} +\mathrm{784c}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{4}} −\mathrm{24c}^{\mathrm{2}} +\mathrm{144}} \\ $$$$\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{7}\right)\:\mathrm{we}\:\mathrm{obtain}: \\ $$$$\frac{\mathrm{8c}\left(\mathrm{8c}^{\mathrm{2}} −\mathrm{28c}\right)}{\mathrm{c}^{\mathrm{2}} −\mathrm{12}}−\frac{\mathrm{c}\left(\mathrm{64c}^{\mathrm{4}} −\mathrm{448c}^{\mathrm{3}} +\mathrm{784c}^{\mathrm{2}} \right)}{\mathrm{c}^{\mathrm{4}} −\mathrm{24c}^{\mathrm{2}} +\mathrm{144}}+\mathrm{c}^{\mathrm{2}} +\mathrm{12}=\mathrm{23c} \\ $$$$\Leftrightarrow\mathrm{c}^{\mathrm{6}} −\mathrm{12c}^{\mathrm{4}} −\mathrm{144c}^{\mathrm{2}} +\mathrm{1728}+\mathrm{64c}^{\mathrm{5}} −\mathrm{224c}^{\mathrm{4}} \\ $$$$−\mathrm{768c}^{\mathrm{3}} +\mathrm{2688c}^{\mathrm{2}} −\mathrm{64c}^{\mathrm{5}} +\mathrm{448c}^{\mathrm{4}} −\mathrm{784c}^{\mathrm{3}} \\ $$$$=\mathrm{23c}^{\mathrm{5}} −\mathrm{552c}^{\mathrm{3}} +\mathrm{3312c} \\ $$$$\Leftrightarrow\mathrm{c}^{\mathrm{6}} −\mathrm{23c}^{\mathrm{5}} +\mathrm{212c}^{\mathrm{4}} −\mathrm{1000c}^{\mathrm{3}} +\mathrm{2544c}^{\mathrm{2}} −\mathrm{3312c}+\mathrm{1728} \\ $$$$=\left(\mathrm{c}−\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{c}−\mathrm{3}\right)\left(\mathrm{c}−\mathrm{4}\right)\left(\mathrm{c}−\mathrm{6}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{c}\in\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{6}\right\}\Rightarrow\mathrm{d}\in\left\{\mathrm{6},\mathrm{4},\mathrm{3},\mathrm{2}\right\} \\ $$$$\Rightarrow\mathrm{a}\in\left\{\mathrm{3},\mathrm{4},\mathrm{4},\mathrm{5}\right\},\mathrm{b}\in\left\{\mathrm{5},\mathrm{4},\mathrm{4},\mathrm{3}\right\} \\ $$$$\Rightarrow\left(\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\right)\in\left\{\left(\mathrm{3},\mathrm{5},\mathrm{2},\mathrm{6}\right),\left(\mathrm{4},\mathrm{4},\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{4},\mathrm{4},\mathrm{3}\right),\left(\mathrm{5},\mathrm{3},\mathrm{6},\mathrm{2}\right)\right\} \\ $$$$\mathrm{We}\:\mathrm{find}\:\mathrm{out}\:\mathrm{two}\:\mathrm{different}\:\mathrm{values}\:\mathrm{of} \\ $$$$\mathrm{S}=\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} \mathrm{being} \\ $$$$\mathrm{S}=\mathrm{4}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{57} \\ $$$$\mathrm{and}\:\mathrm{S}=\mathrm{5}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} =\mathrm{74} \\ $$
Commented by Her_Majesty last updated on 09/Aug/20
you are right... but I don′t fully understand where/why I′m losing 2 solutions
$${you}\:{are}\:{right}…\:{but}\:{I}\:{don}'{t}\:{fully}\:{understand} \\ $$$${where}/{why}\:{I}'{m}\:{losing}\:\mathrm{2}\:{solutions} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *