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Question Number 27559 by Rasheed.Sindhi last updated on 09/Jan/18

Change in Q#27507 Solve simultaneously: 2(√x)+y=13 x+2(√y)=10

$$\mathrm{Change}\:\mathrm{in}\:\mathrm{Q}#\mathrm{27507} \\ $$$$\mathrm{Solve}\:\mathrm{simultaneously}: \\ $$$$\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{y}=\mathrm{13} \\ $$$$\mathrm{x}+\mathrm{2}\sqrt{\mathrm{y}}=\mathrm{10} \\ $$

Commented by Rasheed.Sindhi last updated on 09/Jan/18

x=4,y=9 is solution Method of solving is required.

$$\mathrm{x}=\mathrm{4},\mathrm{y}=\mathrm{9}\:\mathrm{is}\:\mathrm{solution}\: \\ $$$$\mathrm{Method}\:\mathrm{of}\:\mathrm{solving}\:\mathrm{is}\:\mathrm{required}. \\ $$

Answered by Amstrongmazoka last updated on 09/Jan/18

From equation (1) y=13−2(√x) ⇒ substituting for y in equation(2), gives, x+2(√((13−2(√x))))=10 ∴ (√((13−2(√x))))=5−(x/2) Now squaring above sides, gives 13−2(√x)=(5−(x/2))^2 ⇒13−2(√x)=25−5x+(x^2 /4) ∴ −2(√x)=12−5x+(x^2 /4). Again, squaring both sides gives 4x=144−120x+25x^2 +6x^2 −((5x^3 )/2)+(x^4 /(16)). Grouping like terms gives (x^4 /(16))−((5x^3 )/2)+31x^2 −124x+144=0. Multiplying through by 16 gives x^4 −40x^3 +496x^2 −1984x+2304=0. This is a polynomial of degree 4 and it can be solved using factorisation or Newton′ method. Let f(x)=x^4 −40x^3 +496x^2 −1984x+2304 ⇒f(4)=4^4 −40(4^3 )+496(4^2 )−1984(4)+2304=0 By the factor theorem since f(4)=0, ⇒(x−4) is a factor of f(x). The other factors are found as follows: x^3 −36x^2 +352x−576 (x−4)(√(x^4 −40x^3 +496x^2 −1984x+2304)) −(x^4 −4x^3 ) −36x^3 +496x^2 −(−36x^3 +144x^2 ) 352x^2 −1984x −(352x^2 −1408x) −576x+2304 −(−576x+2304) −−−−−−− ∴f(x)=(x−4)(x^3 −36x^2 +352x−576)⇒x=4, x=2.0365, x=19.337 But y=13−2(√x),⇒when x=4, y=13−2(√4)=9, when x=2.0365, y=13−2(√(2.0365))=10.1459 and when x=19.337, y=13−2(√(19.337))=4.2052 thus (x=4, y=9), (x=2.0365, y=10.1459) and (x=19.337, y=4.2052) are the possible pairs from which a solution can be found. amount them, only the first pair satisfy the equations simultaneously. ∴ x=4, y=9

$${From}\:{equation}\:\left(\mathrm{1}\right)\:{y}=\mathrm{13}−\mathrm{2}\sqrt{{x}} \\ $$$$\Rightarrow\:{substituting}\:{for}\:{y}\:{in}\:{equation}\left(\mathrm{2}\right),\:{gives}, \\ $$$${x}+\mathrm{2}\sqrt{\left(\mathrm{13}−\mathrm{2}\sqrt{{x}}\right)}=\mathrm{10}\:\:\:\:\therefore\:\sqrt{\left(\mathrm{13}−\mathrm{2}\sqrt{{x}}\right)}=\mathrm{5}−\frac{{x}}{\mathrm{2}} \\ $$$${Now}\:{squaring}\:{above}\:{sides},\:{gives} \\ $$$$\mathrm{13}−\mathrm{2}\sqrt{{x}}=\left(\mathrm{5}−\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \:\:\:\:\:\:\Rightarrow\mathrm{13}−\mathrm{2}\sqrt{{x}}=\mathrm{25}−\mathrm{5}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\therefore\:−\mathrm{2}\sqrt{{x}}=\mathrm{12}−\mathrm{5}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}.\:\:\:{Again},\:{squaring}\:{both}\:{sides}\:{gives} \\ $$$$\mathrm{4}{x}=\mathrm{144}−\mathrm{120}{x}+\mathrm{25}{x}^{\mathrm{2}} +\mathrm{6}{x}^{\mathrm{2}} −\frac{\mathrm{5}{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{16}}.\: \\ $$$${Grouping}\:{like}\:{terms}\:{gives} \\ $$$$\frac{{x}^{\mathrm{4}} }{\mathrm{16}}−\frac{\mathrm{5}{x}^{\mathrm{3}} }{\mathrm{2}}+\mathrm{31}{x}^{\mathrm{2}} −\mathrm{124}{x}+\mathrm{144}=\mathrm{0}.\:{Multiplying}\:{through}\:{by}\:\mathrm{16}\:{gives} \\ $$$${x}^{\mathrm{4}} −\mathrm{40}{x}^{\mathrm{3}} +\mathrm{496}{x}^{\mathrm{2}} −\mathrm{1984}{x}+\mathrm{2304}=\mathrm{0}.\:{This}\:{is}\:{a}\:{polynomial}\:{of}\:{degree}\:\mathrm{4}\:{and}\:{it}\:{can}\:{be}\:{solved}\:{using}\:{factorisation}\:{or}\:{Newton}'\:{method}. \\ $$$${Let}\:{f}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{40}{x}^{\mathrm{3}} +\mathrm{496}{x}^{\mathrm{2}} −\mathrm{1984}{x}+\mathrm{2304} \\ $$$$\Rightarrow{f}\left(\mathrm{4}\right)=\mathrm{4}^{\mathrm{4}} −\mathrm{40}\left(\mathrm{4}^{\mathrm{3}} \right)+\mathrm{496}\left(\mathrm{4}^{\mathrm{2}} \right)−\mathrm{1984}\left(\mathrm{4}\right)+\mathrm{2304}=\mathrm{0} \\ $$$${By}\:{the}\:{factor}\:{theorem}\:{since}\:{f}\left(\mathrm{4}\right)=\mathrm{0},\:\Rightarrow\left({x}−\mathrm{4}\right)\:{is}\:{a}\:{factor}\:{of}\:{f}\left({x}\right). \\ $$$${The}\:{other}\:{factors}\:{are}\:{found}\:{as}\:{follows}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} \:−\mathrm{36}{x}^{\mathrm{2}} \:+\mathrm{352}{x}−\mathrm{576} \\ $$$$\left({x}−\mathrm{4}\right)\sqrt{{x}^{\mathrm{4}} −\mathrm{40}{x}^{\mathrm{3}} +\mathrm{496}{x}^{\mathrm{2}} −\mathrm{1984}{x}+\mathrm{2304}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left({x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{36}{x}^{\mathrm{3}} +\mathrm{496}{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(−\mathrm{36}{x}^{\mathrm{3}} +\mathrm{144}{x}^{\mathrm{2}} \right)\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{352}{x}^{\mathrm{2}} −\mathrm{1984}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\mathrm{352}{x}^{\mathrm{2}} −\mathrm{1408}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{576}{x}+\mathrm{2304} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(−\mathrm{576}{x}+\mathrm{2304}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−− \\ $$$$\therefore{f}\left({x}\right)=\left({x}−\mathrm{4}\right)\left({x}^{\mathrm{3}} −\mathrm{36}{x}^{\mathrm{2}} +\mathrm{352}{x}−\mathrm{576}\right)\Rightarrow{x}=\mathrm{4},\:\:{x}=\mathrm{2}.\mathrm{0365},\:\:\:{x}=\mathrm{19}.\mathrm{337} \\ $$$${But}\:{y}=\mathrm{13}−\mathrm{2}\sqrt{{x}},\Rightarrow{when}\:{x}=\mathrm{4},\:{y}=\mathrm{13}−\mathrm{2}\sqrt{\mathrm{4}}=\mathrm{9},\:{when}\:{x}=\mathrm{2}.\mathrm{0365},\:\:{y}=\mathrm{13}−\mathrm{2}\sqrt{\mathrm{2}.\mathrm{0365}}=\mathrm{10}.\mathrm{1459}\:{and}\:{when}\:{x}=\mathrm{19}.\mathrm{337},\:{y}=\mathrm{13}−\mathrm{2}\sqrt{\mathrm{19}.\mathrm{337}}=\mathrm{4}.\mathrm{2052} \\ $$$${thus}\:\left({x}=\mathrm{4},\:{y}=\mathrm{9}\right),\:\:\:\left({x}=\mathrm{2}.\mathrm{0365},\:{y}=\mathrm{10}.\mathrm{1459}\right)\:\:{and}\:\left({x}=\mathrm{19}.\mathrm{337},\:{y}=\mathrm{4}.\mathrm{2052}\right) \\ $$$${are}\:{the}\:{possible}\:{pairs}\:{from}\:{which}\:{a}\:{solution}\:{can}\:{be}\:{found}. \\ $$$${amount}\:{them},\:{only}\:{the}\:{first}\:{pair}\:{satisfy}\:{the}\:{equations}\:{simultaneously}. \\ $$$$\therefore\:{x}=\mathrm{4},\:{y}=\mathrm{9} \\ $$

Commented by Rasheed.Sindhi last updated on 09/Jan/18

In finding factors trial method is involved. AnyWay ThanX-a-lot!

$$\mathrm{In}\:\mathrm{finding}\:\mathrm{factors}\:\mathrm{trial}\:\mathrm{method}\: \\ $$$$\mathrm{is}\:\mathrm{involved}. \\ $$$$\mathcal{A}{nyWay}\:\mathcal{T}{han}\mathcal{X}-{a}-{lot}! \\ $$

Commented by Rasheed.Sindhi last updated on 09/Jan/18

What if we are only sure of real roots?

$$\mathrm{What}\:\mathrm{if}\:\mathrm{we}\:\mathrm{are}\:\mathrm{only}\:\mathrm{sure}\:\mathrm{of} \\ $$$$\boldsymbol{\mathrm{real}}\:\mathrm{roots}? \\ $$

Answered by Rasheed.Sindhi last updated on 10/Jan/18

Trying to find solution in whole numbers. 2(√x)+y=13........(i) x+2(√y)=10........(ii) (i)⇒(√x)=((13−y)/2)..........(iii) (ii)⇒x=10−2(√y) (√x)=(√(10−2(√y)))......(iv) (iii) & (iv)⇒((13−y)/2)=(√(10−2(√y))) (√(p−q(√c))) can be changed into a+b(√c) form in some cases and the process is as under. Let (√(10−2(√y)))=a+b(√y) a,b∈Q ((√(10−2(√y))))^2 =(a+b(√y))^2 a^2 +b^2 y+2ab(√y)=10−2(√y) By comparing the coefficients of (√y) and the terms not containing (√y) we′ll get: a^2 +b^2 y=10 ∧ 2ab=−2 ab=−1 b=−1/a a^2 +(−1/a)^2 y=10 a^4 −10a^2 +y=0 a^2 =((10±(√(100−4y)))/2) ((10±(√(100−4y)))/2) is perfect square rational number⇒100−4y is perfect square rational number and non-negative. ∴ y=0,9,16,21 y=0⇒a=((10±10)/2)=(√(10))^(×) ,0 a≠0 because in that case b=−1/0=∞ y=9⇒a=((10±8)/2)=3,1 y=16⇒a=((10±6)/2)=(√8)^(×) ,(√2)^(×) y=21⇒a=((10±4)/2)=(√7)^(×) ,(√3)^(×) So y may be 9 and a=1,3 a=3,b=−1/3 & y=9 satisfy the following (√(10−2(√y)))=a+b(√y) (√(10−2(√9)))=3−(1/3)(√9) 2=2 Hence y=9 Continue

$$\mathrm{Trying}\:\mathrm{to}\:\mathrm{find}\:\mathrm{solution}\:\mathrm{in} \\ $$$$\mathrm{whole}\:\mathrm{numbers}. \\ $$$$\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{y}=\mathrm{13}........\left(\mathrm{i}\right) \\ $$$$\mathrm{x}+\mathrm{2}\sqrt{\mathrm{y}}=\mathrm{10}........\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)\Rightarrow\sqrt{\mathrm{x}}=\frac{\mathrm{13}−\mathrm{y}}{\mathrm{2}}..........\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{ii}\right)\Rightarrow\mathrm{x}=\mathrm{10}−\mathrm{2}\sqrt{\mathrm{y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{x}}=\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{y}}}......\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{iii}\right)\:\&\:\left(\mathrm{iv}\right)\Rightarrow\frac{\mathrm{13}−\mathrm{y}}{\mathrm{2}}=\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{y}}} \\ $$$$\sqrt{\mathrm{p}−\mathrm{q}\sqrt{\mathrm{c}}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{changed}\:\mathrm{into} \\ $$$$\mathrm{a}+\mathrm{b}\sqrt{\mathrm{c}}\:\mathrm{form}\:\mathrm{in}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{process}\:\mathrm{is}\:\mathrm{as}\:\mathrm{under}. \\ $$$$\mathrm{Let}\:\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{y}}}=\mathrm{a}+\mathrm{b}\sqrt{\mathrm{y}}\:\:\:\:\mathrm{a},\mathrm{b}\in\mathbb{Q} \\ $$$$\:\:\:\:\:\left(\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{y}}}\right)^{\mathrm{2}} =\left(\mathrm{a}+\mathrm{b}\sqrt{\mathrm{y}}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \mathrm{y}+\mathrm{2ab}\sqrt{\mathrm{y}}=\mathrm{10}−\mathrm{2}\sqrt{\mathrm{y}} \\ $$$$\:\:\mathrm{By}\:\mathrm{comparing}\:\mathrm{the}\:\mathrm{coefficients} \\ $$$$\mathrm{of}\:\sqrt{\mathrm{y}}\:\mathrm{and}\:\mathrm{the}\:\mathrm{terms}\:\mathrm{not}\:\mathrm{containing} \\ $$$$\sqrt{\mathrm{y}}\:\mathrm{we}'\mathrm{ll}\:\mathrm{get}: \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \mathrm{y}=\mathrm{10}\:\:\:\wedge\:\mathrm{2ab}=−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ab}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}=−\mathrm{1}/\mathrm{a} \\ $$$$\mathrm{a}^{\mathrm{2}} +\left(−\mathrm{1}/\mathrm{a}\right)^{\mathrm{2}} \mathrm{y}=\mathrm{10} \\ $$$$\mathrm{a}^{\mathrm{4}} −\mathrm{10a}^{\mathrm{2}} +\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{2}} =\frac{\mathrm{10}\pm\sqrt{\mathrm{100}−\mathrm{4y}}}{\mathrm{2}} \\ $$$$\:\:\frac{\mathrm{10}\pm\sqrt{\mathrm{100}−\mathrm{4y}}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\mathrm{rational}\:\mathrm{number}\Rightarrow\mathrm{100}−\mathrm{4y}\:\mathrm{is} \\ $$$$\mathrm{perfect}\:\mathrm{square}\:\mathrm{rational}\:\mathrm{number} \\ $$$$\mathrm{and}\:\mathrm{non}-\mathrm{negative}. \\ $$$$\therefore\:\mathrm{y}=\mathrm{0},\mathrm{9},\mathrm{16},\mathrm{21} \\ $$$$\mathrm{y}=\mathrm{0}\Rightarrow\mathrm{a}=\frac{\mathrm{10}\pm\mathrm{10}}{\mathrm{2}}=\overset{×} {\sqrt{\mathrm{10}}},\mathrm{0} \\ $$$$\mathrm{a}\neq\mathrm{0}\:\mathrm{because}\:\mathrm{in}\:\mathrm{that}\:\mathrm{case}\:\mathrm{b}=−\mathrm{1}/\mathrm{0}=\infty \\ $$$$\mathrm{y}=\mathrm{9}\Rightarrow\mathrm{a}=\frac{\mathrm{10}\pm\mathrm{8}}{\mathrm{2}}=\mathrm{3},\mathrm{1} \\ $$$$\mathrm{y}=\mathrm{16}\Rightarrow\mathrm{a}=\frac{\mathrm{10}\pm\mathrm{6}}{\mathrm{2}}=\overset{×} {\sqrt{\mathrm{8}}}\:,\overset{×} {\sqrt{\mathrm{2}}}\: \\ $$$$\mathrm{y}=\mathrm{21}\Rightarrow\mathrm{a}=\frac{\mathrm{10}\pm\mathrm{4}}{\mathrm{2}}=\overset{×} {\sqrt{\mathrm{7}}}\:,\overset{×} {\sqrt{\mathrm{3}}} \\ $$$$\mathrm{So}\:\mathrm{y}\:\mathrm{may}\:\mathrm{be}\:\mathrm{9}\:\:\mathrm{and}\:\mathrm{a}=\mathrm{1},\mathrm{3} \\ $$$$\mathrm{a}=\mathrm{3},\mathrm{b}=−\mathrm{1}/\mathrm{3}\:\&\:\mathrm{y}=\mathrm{9}\:\mathrm{satisfy}\:\mathrm{the} \\ $$$$\mathrm{following} \\ $$$$\:\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{y}}}=\mathrm{a}+\mathrm{b}\sqrt{\mathrm{y}}\: \\ $$$$\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{9}}}=\mathrm{3}−\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{9}} \\ $$$$\:\:\:\:\mathrm{2}=\mathrm{2} \\ $$$$\mathrm{Hence}\:\mathrm{y}=\mathrm{9} \\ $$$$\mathrm{Continue} \\ $$

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