Menu Close

Question-203202

Tinku Tara 0 Comments
Pin




Question Number 203202 by Abdullahrussell last updated on 12/Jan/24
Answered by cortano12 last updated on 12/Jan/24
≤
$$\:\:\:\underbrace{\leq} \\ $$
Answered by mr W last updated on 12/Jan/24
abc+d(ab+bc+ca)=14 ⇒abc=14+4d abcd=(14+4d)d=30 ⇒2d^2 +7d−15=0 ⇒(2d−3)(d+5)=0 ⇒d=(3/2) or −5 abc=((30)/(3/2))=20 or ((30)/(−5))=−6 ab+(a+b)c=−4 abc−3c^2 =−4c 3c^2 −4c−abc=0 ⇒3c^2 −4c−20=0 or 3c^2 −4c+6=0 (rejected) ⇒(3c−10)(c+2)=0 ⇒c=((10)/3) or −2 ab=−4+3c=6 (rejected) or −10 a+b=−3 ⇒(a, b)=(−5, 2) or (2, −5) (a, b, c, d)=(−5, 2, −2, (3/2)) or (2, −5, −2, (3/2)) a^2 +b^2 +c^2 +d^2 =25+4+4+(9/4)=((141)/4) ✓
$${abc}+{d}\left({ab}+{bc}+{ca}\right)=\mathrm{14} \\ $$$$\Rightarrow{abc}=\mathrm{14}+\mathrm{4}{d} \\ $$$${abcd}=\left(\mathrm{14}+\mathrm{4}{d}\right){d}=\mathrm{30} \\ $$$$\Rightarrow\mathrm{2}{d}^{\mathrm{2}} +\mathrm{7}{d}−\mathrm{15}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{d}−\mathrm{3}\right)\left({d}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow{d}=\frac{\mathrm{3}}{\mathrm{2}}\:{or}\:−\mathrm{5}\: \\ $$$${abc}=\frac{\mathrm{30}}{\frac{\mathrm{3}}{\mathrm{2}}}=\mathrm{20}\:{or}\:\frac{\mathrm{30}}{−\mathrm{5}}=−\mathrm{6} \\ $$$${ab}+\left({a}+{b}\right){c}=−\mathrm{4} \\ $$$${abc}−\mathrm{3}{c}^{\mathrm{2}} =−\mathrm{4}{c} \\ $$$$\mathrm{3}{c}^{\mathrm{2}} −\mathrm{4}{c}−{abc}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{c}^{\mathrm{2}} −\mathrm{4}{c}−\mathrm{20}=\mathrm{0}\:{or}\:\mathrm{3}{c}^{\mathrm{2}} −\mathrm{4}{c}+\mathrm{6}=\mathrm{0}\:\left({rejected}\right) \\ $$$$\Rightarrow\left(\mathrm{3}{c}−\mathrm{10}\right)\left({c}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{c}=\frac{\mathrm{10}}{\mathrm{3}}\:{or}\:−\mathrm{2} \\ $$$${ab}=−\mathrm{4}+\mathrm{3}{c}=\mathrm{6}\:\left({rejected}\right)\:{or}\:−\mathrm{10} \\ $$$${a}+{b}=−\mathrm{3} \\ $$$$\Rightarrow\left({a},\:{b}\right)=\left(−\mathrm{5},\:\mathrm{2}\right)\:{or}\:\left(\mathrm{2},\:−\mathrm{5}\right) \\ $$$$\left({a},\:{b},\:{c},\:{d}\right)=\left(−\mathrm{5},\:\mathrm{2},\:−\mathrm{2},\:\frac{\mathrm{3}}{\mathrm{2}}\right)\:{or}\:\left(\mathrm{2},\:−\mathrm{5},\:−\mathrm{2},\:\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{25}+\mathrm{4}+\mathrm{4}+\frac{\mathrm{9}}{\mathrm{4}}=\frac{\mathrm{141}}{\mathrm{4}}\:\checkmark \\ $$
Answered by ajfour last updated on 12/Jan/24
ab=m c=((4+m)/3) d=((30×3)/(m(4+m))) ((m(4+m))/3)+(m−3c)d=14 ((m(4+m))/3)−((90×4)/(m(4+m)))=14 ⇒m^2 (((4+m)/3))^2 =((14m(4+m))/3)+120 t^2 −14−120=0 (t−7)^2 =13^2 t=20, −6 m(4+m)=60, −18 (m+2)^2 =64, −14 lets take m=6, −10 & 9−2m+(((4+m)/3))^2 +((8100)/(m^2 (4+m)^2 ))=?=Q with m=6 Q_1 =−3+((100)/9)+((81)/(36))=((481−108)/(36)) Q_1 =((373)/(36)) & with m=−10 Q_2 =29+4+((81)/(36))=35+(1/4)=((141)/4)
$${ab}={m} \\ $$$${c}=\frac{\mathrm{4}+{m}}{\mathrm{3}} \\ $$$${d}=\frac{\mathrm{30}×\mathrm{3}}{{m}\left(\mathrm{4}+{m}\right)} \\ $$$$\frac{{m}\left(\mathrm{4}+{m}\right)}{\mathrm{3}}+\left({m}−\mathrm{3}{c}\right){d}=\mathrm{14} \\ $$$$\frac{{m}\left(\mathrm{4}+{m}\right)}{\mathrm{3}}−\frac{\mathrm{90}×\mathrm{4}}{{m}\left(\mathrm{4}+{m}\right)}=\mathrm{14} \\ $$$$\Rightarrow{m}^{\mathrm{2}} \:\left(\frac{\mathrm{4}+{m}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{14}{m}\left(\mathrm{4}+{m}\right)}{\mathrm{3}}+\mathrm{120} \\ $$$${t}^{\mathrm{2}} −\mathrm{14}−\mathrm{120}=\mathrm{0} \\ $$$$\left({t}−\mathrm{7}\right)^{\mathrm{2}} =\mathrm{13}^{\mathrm{2}} \\ $$$${t}=\mathrm{20},\:−\mathrm{6} \\ $$$${m}\left(\mathrm{4}+{m}\right)=\mathrm{60},\:−\mathrm{18} \\ $$$$\left({m}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{64},\:−\mathrm{14} \\ $$$${lets}\:{take}\:{m}=\mathrm{6},\:−\mathrm{10} \\ $$$$\& \\ $$$$\mathrm{9}−\mathrm{2}{m}+\left(\frac{\mathrm{4}+{m}}{\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{8100}}{{m}^{\mathrm{2}} \left(\mathrm{4}+{m}\right)^{\mathrm{2}} }=?={Q} \\ $$$${with}\:{m}=\mathrm{6} \\ $$$${Q}_{\mathrm{1}} =−\mathrm{3}+\frac{\mathrm{100}}{\mathrm{9}}+\frac{\mathrm{81}}{\mathrm{36}}=\frac{\mathrm{481}−\mathrm{108}}{\mathrm{36}} \\ $$$${Q}_{\mathrm{1}} =\frac{\mathrm{373}}{\mathrm{36}} \\ $$$$\&\:\:{with}\:\:{m}=−\mathrm{10} \\ $$$${Q}_{\mathrm{2}} =\mathrm{29}+\mathrm{4}+\frac{\mathrm{81}}{\mathrm{36}}=\mathrm{35}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{141}}{\mathrm{4}} \\ $$

Pin

Leave a Reply

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