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Question Number 41984 by Tawa1 last updated on 16/Aug/18
solve simultaneously: 2(√k) + h = 9 ....... (i) k + 2(√h) = 3 ....... (ii)
$$\mathrm{solve}\:\mathrm{simultaneously}:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\sqrt{\mathrm{k}}\:\:+\:\mathrm{h}\:=\:\mathrm{9}\:\:\:…….\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{k}\:+\:\mathrm{2}\sqrt{\mathrm{h}}\:\:=\:\mathrm{3}\:\:…….\:\left(\mathrm{ii}\right) \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
x+2(√y) −9=0 2(√x) +y−3=0 2(√y) =(9−x) 4y=(9−x)^2 (9−x)^2 =4×1×y
$${x}+\mathrm{2}\sqrt{{y}}\:−\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{2}\sqrt{{x}}\:+{y}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}\sqrt{{y}}\:=\left(\mathrm{9}−{x}\right) \\ $$$$\mathrm{4}{y}=\left(\mathrm{9}−{x}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{9}−{x}\right)^{\mathrm{2}} =\mathrm{4}×\mathrm{1}×{y} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
this is a parabola eqn 2(√x) =(y−3) 4x=(y−3)^2 (y−3)^2 =4x this is also a parabola
$${this}\:{is}\:{a}\:{parabola}\:{eqn}\: \\ $$$$\mathrm{2}\sqrt{{x}}\:=\left({y}−\mathrm{3}\right) \\ $$$$\mathrm{4}{x}=\left({y}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{4}{x} \\ $$$${this}\:{is}\:{also}\:{a}\:{parabola} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
no solution...
$${no}\:{solution}… \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
if converted to parabola by squaring...then solution x=3.8(approx) y=7.5 approx h=3.8 k=7.5
$${if}\:{converted}\:{to}\:{parabola}\:{by}\:{squaring}…{then}\:{solution} \\ $$$${x}=\mathrm{3}.\mathrm{8}\left({approx}\right)\:\:\:{y}=\mathrm{7}.\mathrm{5}\:{approx} \\ $$$${h}=\mathrm{3}.\mathrm{8}\:\:\:\:{k}=\mathrm{7}.\mathrm{5}\:\:\: \\ $$
Commented by MJS last updated on 16/Aug/18
in this case we have 2 real and 2 complex points A= (((3.76)),((6.88)) ) B= (((15.60)),((10.90)) ) C= (((8.32+3.47i)),((−2.89−1.18i)) ) D= (((8.32−3.47i)),((−2.89+1.18i)) )
$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{have}\:\mathrm{2}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{complex} \\ $$$$\mathrm{points} \\ $$$${A}=\begin{pmatrix}{\mathrm{3}.\mathrm{76}}\\{\mathrm{6}.\mathrm{88}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{\mathrm{15}.\mathrm{60}}\\{\mathrm{10}.\mathrm{90}}\end{pmatrix} \\ $$$${C}=\begin{pmatrix}{\mathrm{8}.\mathrm{32}+\mathrm{3}.\mathrm{47i}}\\{−\mathrm{2}.\mathrm{89}−\mathrm{1}.\mathrm{18i}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{8}.\mathrm{32}−\mathrm{3}.\mathrm{47i}}\\{−\mathrm{2}.\mathrm{89}+\mathrm{1}.\mathrm{18i}}\end{pmatrix} \\ $$
Answered by MJS last updated on 16/Aug/18
(i) (√k)=((9−h)/2) ⇒ 9−h≥0 ⇔ h≤9 k=(((9−h)^2 )/4) (ii) k=3−2(√h) ⇒ h≥0 k=(((9−h)^2 )/4) ∧ 0≤h≤9 ⇒ 0≤k≤((81)/4) k=3−2(√h) ∧ 0≤h≤9 ⇒ −3≤h≤3 ⇒ 0≤k≤3 (i) h=9−2(√k) ∧ 0≤k≤3 ⇒ (9−2(√3))≤h≤9 (ii) (√h)=((3−k)/2) ⇒ 3−k≥0 ⇔ k≤3 h=(((3−k)^2 )/4) ∧ 0≤k≤3 ⇒ 0≤h≤(9/4) so we have 2 musts: (9−2(√3))≤h≤9 and 0≤h≤(9/4) 9−2(√3)≈5.54 (9/4)=2.25 h≥5.54 and h≤2.25 is impossible ⇒ no real solution
$$\left({i}\right) \\ $$$$\sqrt{{k}}=\frac{\mathrm{9}−{h}}{\mathrm{2}}\:\Rightarrow\:\mathrm{9}−{h}\geqslant\mathrm{0}\:\Leftrightarrow\:{h}\leqslant\mathrm{9} \\ $$$${k}=\frac{\left(\mathrm{9}−{h}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\left({ii}\right) \\ $$$${k}=\mathrm{3}−\mathrm{2}\sqrt{{h}}\:\Rightarrow\:{h}\geqslant\mathrm{0} \\ $$$$ \\ $$$${k}=\frac{\left(\mathrm{9}−{h}\right)^{\mathrm{2}} }{\mathrm{4}}\:\wedge\:\mathrm{0}\leqslant{h}\leqslant\mathrm{9}\:\Rightarrow\:\mathrm{0}\leqslant{k}\leqslant\frac{\mathrm{81}}{\mathrm{4}} \\ $$$${k}=\mathrm{3}−\mathrm{2}\sqrt{{h}}\:\wedge\:\mathrm{0}\leqslant{h}\leqslant\mathrm{9}\:\Rightarrow\:−\mathrm{3}\leqslant{h}\leqslant\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{0}\leqslant{k}\leqslant\mathrm{3} \\ $$$$\left({i}\right) \\ $$$${h}=\mathrm{9}−\mathrm{2}\sqrt{{k}}\:\wedge\:\mathrm{0}\leqslant{k}\leqslant\mathrm{3}\:\Rightarrow\:\left(\mathrm{9}−\mathrm{2}\sqrt{\mathrm{3}}\right)\leqslant{h}\leqslant\mathrm{9} \\ $$$$\left({ii}\right) \\ $$$$\sqrt{{h}}=\frac{\mathrm{3}−{k}}{\mathrm{2}}\:\Rightarrow\:\mathrm{3}−{k}\geqslant\mathrm{0}\:\Leftrightarrow\:{k}\leqslant\mathrm{3} \\ $$$${h}=\frac{\left(\mathrm{3}−{k}\right)^{\mathrm{2}} }{\mathrm{4}}\:\wedge\:\mathrm{0}\leqslant{k}\leqslant\mathrm{3}\:\Rightarrow\:\mathrm{0}\leqslant{h}\leqslant\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{2}\:\mathrm{musts}: \\ $$$$\left(\mathrm{9}−\mathrm{2}\sqrt{\mathrm{3}}\right)\leqslant{h}\leqslant\mathrm{9}\:\mathrm{and}\:\mathrm{0}\leqslant{h}\leqslant\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{9}−\mathrm{2}\sqrt{\mathrm{3}}\approx\mathrm{5}.\mathrm{54} \\ $$$$\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{2}.\mathrm{25} \\ $$$${h}\geqslant\mathrm{5}.\mathrm{54}\:\mathrm{and}\:{h}\leqslant\mathrm{2}.\mathrm{25}\:\mathrm{is}\:\mathrm{impossible}\:\Rightarrow\:\mathrm{no}\:\mathrm{real} \\ $$$$\mathrm{solution} \\ $$
Commented by Tawa1 last updated on 16/Aug/18
God bless you sir. am still expecting the polynomial equation sir. any time
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{am}\:\mathrm{still}\:\mathrm{expecting}\:\mathrm{the}\:\mathrm{polynomial}\:\mathrm{equation}\:\mathrm{sir}. \\ $$$$\mathrm{any}\:\mathrm{time} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
(√k) =((9−h)/2) k=((81−18h+h^2 )/4) (√h) =((3−k)/2) h=((9−6k+k^2 )/4) so k=((81−18(((9−6k+k^2 )/4))+((81+36k^2 +k^4 −108k−12k^3 +18k^2 )/(16)))/4) 4k=81×16−18×4(9−6k+k^2 )+81−108k+54k^2 −12k^3 +k^4 4k=1296−18×4×9+18×4×6k−18×4k^2 −108k+54k^2 −12k^3 +k^4 k^4 −12k^3 +k^2 (54−72)+k(−108+18×4×6−4)+1296−18×4×9=0 k^4 −12k^3 −18k^2 +k(−108+432−4)+1296−648=0 k^4 −12k^3 −18k^2 +320k+648=0
$$\sqrt{{k}}\:=\frac{\mathrm{9}−{h}}{\mathrm{2}} \\ $$$${k}=\frac{\mathrm{81}−\mathrm{18}{h}+{h}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\sqrt{{h}}\:=\frac{\mathrm{3}−{k}}{\mathrm{2}} \\ $$$${h}=\frac{\mathrm{9}−\mathrm{6}{k}+{k}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${so}\:{k}=\frac{\mathrm{81}−\mathrm{18}\left(\frac{\mathrm{9}−\mathrm{6}{k}+{k}^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{\mathrm{81}+\mathrm{36}{k}^{\mathrm{2}} +{k}^{\mathrm{4}} −\mathrm{108}{k}−\mathrm{12}{k}^{\mathrm{3}} +\mathrm{18}{k}^{\mathrm{2}} }{\mathrm{16}}}{\mathrm{4}} \\ $$$$\mathrm{4}{k}=\mathrm{81}×\mathrm{16}−\mathrm{18}×\mathrm{4}\left(\mathrm{9}−\mathrm{6}{k}+{k}^{\mathrm{2}} \right)+\mathrm{81}−\mathrm{108}{k}+\mathrm{54}{k}^{\mathrm{2}} −\mathrm{12}{k}^{\mathrm{3}} +{k}^{\mathrm{4}} \\ $$$$\mathrm{4}{k}=\mathrm{1296}−\mathrm{18}×\mathrm{4}×\mathrm{9}+\mathrm{18}×\mathrm{4}×\mathrm{6}{k}−\mathrm{18}×\mathrm{4}{k}^{\mathrm{2}} −\mathrm{108}{k}+\mathrm{54}{k}^{\mathrm{2}} −\mathrm{12}{k}^{\mathrm{3}} +{k}^{\mathrm{4}} \\ $$$${k}^{\mathrm{4}} −\mathrm{12}{k}^{\mathrm{3}} +{k}^{\mathrm{2}} \left(\mathrm{54}−\mathrm{72}\right)+{k}\left(−\mathrm{108}+\mathrm{18}×\mathrm{4}×\mathrm{6}−\mathrm{4}\right)+\mathrm{1296}−\mathrm{18}×\mathrm{4}×\mathrm{9}=\mathrm{0} \\ $$$${k}^{\mathrm{4}} −\mathrm{12}{k}^{\mathrm{3}} −\mathrm{18}{k}^{\mathrm{2}} +{k}\left(−\mathrm{108}+\mathrm{432}−\mathrm{4}\right)+\mathrm{1296}−\mathrm{648}=\mathrm{0} \\ $$$${k}^{\mathrm{4}} −\mathrm{12}{k}^{\mathrm{3}} −\mathrm{18}{k}^{\mathrm{2}} +\mathrm{320}{k}+\mathrm{648}=\mathrm{0} \\ $$
Commented by MJS last updated on 10/Sep/18
typo somewhere. it must be k^4 −12k^3 −18k^2 +260k+729=0 k_1 =6.87588... k_2 =10.9001... but we squared 2 times to get here and both resulting pairs of h, k don′t solve the given equation system
$$\mathrm{typo}\:\mathrm{somewhere}.\:\mathrm{it}\:\mathrm{must}\:\mathrm{be} \\ $$$${k}^{\mathrm{4}} −\mathrm{12}{k}^{\mathrm{3}} −\mathrm{18}{k}^{\mathrm{2}} +\mathrm{260}{k}+\mathrm{729}=\mathrm{0} \\ $$$${k}_{\mathrm{1}} =\mathrm{6}.\mathrm{87588}… \\ $$$${k}_{\mathrm{2}} =\mathrm{10}.\mathrm{9001}… \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{squared}\:\mathrm{2}\:\mathrm{times}\:\mathrm{to}\:\mathrm{get}\:\mathrm{here}\:\mathrm{and}\:\mathrm{both} \\ $$$$\mathrm{resulting}\:\mathrm{pairs}\:\mathrm{of}\:{h},\:{k}\:\mathrm{don}'\mathrm{t}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equation}\:\mathrm{system} \\ $$

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