Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$$\mathrm{typo}\:\mathrm{somewhere}.\:\mathrm{it}\:\mathrm{must}\:\mathrm{be} \\ $$$${k}^{\mathrm{4}} −\mathrm{12}{k}^{\mathrm{3}} −\mathrm{18}{k}^{\mathrm{2}} +\mathrm{260}{k}+\mathrm{729}=\mathrm{0} \\ $$$${k}_{\mathrm{1}} =\mathrm{6}.\mathrm{87588}... \\ $$$${k}_{\mathrm{2}} =\mathrm{10}.\mathrm{9001}... \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{squared}\:\mathrm{2}\:\mathrm{times}\:\mathrm{to}\:\mathrm{get}\:\mathrm{here}\:\mathrm{and}\:\mathrm{both} \\ $$$$\mathrm{resulting}\:\mathrm{pairs}\:\mathrm{of}\:{h},\:{k}\:\mathrm{don}'\mathrm{t}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equation}\:\mathrm{system} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com