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Question Number 61479 by Tawa1 last updated on 03/Jun/19

Solve for x: ((6(√(2x)))/(x − 1)) + ((5(√(x − 1)))/(2x)) = 13

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\:\:\frac{\mathrm{6}\sqrt{\mathrm{2x}}}{\mathrm{x}\:−\:\mathrm{1}}\:+\:\frac{\mathrm{5}\sqrt{\mathrm{x}\:−\:\mathrm{1}}}{\mathrm{2x}}\:\:\:=\:\:\mathrm{13} \\ $$

Commented by MJS last updated on 03/Jun/19

we cannot generally solve this...

$$\mathrm{we}\:\mathrm{cannot}\:\mathrm{generally}\:\mathrm{solve}\:\mathrm{this}... \\ $$

Answered by ajfour last updated on 03/Jun/19

let x=sec^2 θ ⇒ 6(√2)sec θcot^2 θ+((5tan θcos^2 θ)/2)=13 ((12(√2)cos θ)/(sin^2 θ))+5sin θcos θ = 26 (((12(√2)+5sin^3 θ)/(sin^2 θ)))^2 (1−sin^2 θ)=676 ⇒ (a+bt^3 )^2 (1−t^2 )=ct^4 where a=12(√2), b=5, c=676, and t=sin θ (A polynomial of degree 8)

$${let}\:\:{x}=\mathrm{sec}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\mathrm{6}\sqrt{\mathrm{2}}\mathrm{sec}\:\theta\mathrm{cot}\:^{\mathrm{2}} \theta+\frac{\mathrm{5tan}\:\theta\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{2}}=\mathrm{13} \\ $$$$\:\:\:\frac{\mathrm{12}\sqrt{\mathrm{2}}\mathrm{cos}\:\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta}+\mathrm{5sin}\:\theta\mathrm{cos}\:\theta\:=\:\mathrm{26} \\ $$$$\:\left(\frac{\mathrm{12}\sqrt{\mathrm{2}}+\mathrm{5sin}\:^{\mathrm{3}} \theta}{\mathrm{sin}\:^{\mathrm{2}} \theta}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta\right)=\mathrm{676} \\ $$$$\Rightarrow\:\:\left({a}+{bt}^{\mathrm{3}} \right)^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)={ct}^{\mathrm{4}} \\ $$$${where}\:\:{a}=\mathrm{12}\sqrt{\mathrm{2}},\:{b}=\mathrm{5},\:{c}=\mathrm{676}, \\ $$$$\:\:\:\:{and}\:\:{t}=\mathrm{sin}\:\theta \\ $$$$\left({A}\:{polynomial}\:{of}\:{degree}\:\mathrm{8}\right) \\ $$

Commented by Tawa1 last updated on 03/Jun/19

Ohh.

$$\mathrm{Ohh}.\:\: \\ $$

Answered by MJS last updated on 03/Jun/19

x≈2.028522 is the only solution ∈R (found no exact value)

$${x}\approx\mathrm{2}.\mathrm{028522}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\in\mathbb{R} \\ $$$$\left(\mathrm{found}\:\mathrm{no}\:\mathrm{exact}\:\mathrm{value}\right) \\ $$

Commented by Tawa1 last updated on 03/Jun/19

Any workings sir

$$\mathrm{Any}\:\mathrm{workings}\:\mathrm{sir} \\ $$

Commented by MJS last updated on 03/Jun/19

we can square it 2 times but we end up with a polynome of degree 8 which has got no trivial solution [squaring like this: (√a)+(√b)=c a+(√a)(√b)+b=c^2 ab=(c^2 −a−b)^2 ] approximated using a calculator the equation is defined for x>1 so I started with x=1.5, x=2, x=2.5

$$\mathrm{we}\:\mathrm{can}\:\mathrm{square}\:\mathrm{it}\:\mathrm{2}\:\mathrm{times}\:\mathrm{but}\:\mathrm{we}\:\mathrm{end}\:\mathrm{up}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{8}\:\mathrm{which}\:\mathrm{has}\:\mathrm{got}\:\mathrm{no} \\ $$$$\mathrm{trivial}\:\mathrm{solution} \\ $$$$\left[\mathrm{squaring}\:\mathrm{like}\:\mathrm{this}:\right. \\ $$$$\sqrt{{a}}+\sqrt{{b}}={c} \\ $$$${a}+\sqrt{{a}}\sqrt{{b}}+{b}={c}^{\mathrm{2}} \\ $$$$\left.{ab}=\left({c}^{\mathrm{2}} −{a}−{b}\right)^{\mathrm{2}} \right] \\ $$$$ \\ $$$$\mathrm{approximated}\:\mathrm{using}\:\mathrm{a}\:\mathrm{calculator} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}>\mathrm{1}\:\mathrm{so}\:\mathrm{I}\:\mathrm{started} \\ $$$$\mathrm{with}\:{x}=\mathrm{1}.\mathrm{5},\:{x}=\mathrm{2},\:{x}=\mathrm{2}.\mathrm{5} \\ $$

Answered by behi83417@gmail.com last updated on 03/Jun/19

(√(2x))=a,(√(x−1))=b⇒x=(a^2 /2)=b^2 +1 ((6a)/b^2 )+((5b)/a^2 )=13,a^2 =2b^2 +2 ⇒ { ((6a^3 +5b^3 =13a^2 b^2 )),((a^2 −2b^2 =2⇒a^3 =2a(b^2 +1))) :} ⇒ { ((5(a+b)(a^2 −ab+b^2 )+a^3 =13a^2 b^2 )),(((a−b)(a+b)=b^2 +2)) :} ⇒_(ab=q) ^(a+b=p) { ((5p(p^2 −3q)=13a^2 b^2 −2a(b^2 +1))),((p^2 (p^2 −4q)=(b^2 +2)^2 )) :} ⇒ { ((5p^3 −15pq=13a^2 ((a^2 /2)−1)−2a((a^2 /2)))),((p^4 −4p^2 q=((a^2 /2)+1)^2 )) :} ⇒ { ((80p^3 −240pq=8p(13a^4 −2a^3 −26a^2 ))),((60p^4 −240p^2 q=15(a^4 +4a^2 +4))) :} 140p^4 −8pa^2 (13a^2 −2a−26)−15(a^2 +2)^2 =0

$$\sqrt{\mathrm{2x}}=\mathrm{a},\sqrt{\mathrm{x}−\mathrm{1}}=\mathrm{b}\Rightarrow\mathrm{x}=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{b}^{\mathrm{2}} +\mathrm{1} \\ $$$$\frac{\mathrm{6a}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{5b}}{\mathrm{a}^{\mathrm{2}} }=\mathrm{13},\mathrm{a}^{\mathrm{2}} =\mathrm{2b}^{\mathrm{2}} +\mathrm{2} \\ $$$$\Rightarrow\begin{cases}{\mathrm{6a}^{\mathrm{3}} +\mathrm{5b}^{\mathrm{3}} =\mathrm{13a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} }\\{\mathrm{a}^{\mathrm{2}} −\mathrm{2b}^{\mathrm{2}} =\mathrm{2}\Rightarrow\mathrm{a}^{\mathrm{3}} =\mathrm{2a}\left(\mathrm{b}^{\mathrm{2}} +\mathrm{1}\right)}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{5}\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} \right)+\mathrm{a}^{\mathrm{3}} =\mathrm{13a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} }\\{\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{b}^{\mathrm{2}} +\mathrm{2}}\end{cases} \\ $$$$\underset{\mathrm{ab}=\mathrm{q}} {\overset{\mathrm{a}+\mathrm{b}=\mathrm{p}} {\Rightarrow}}\begin{cases}{\mathrm{5p}\left(\mathrm{p}^{\mathrm{2}} −\mathrm{3q}\right)=\mathrm{13a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} −\mathrm{2a}\left(\mathrm{b}^{\mathrm{2}} +\mathrm{1}\right)}\\{\mathrm{p}^{\mathrm{2}} \left(\mathrm{p}^{\mathrm{2}} −\mathrm{4q}\right)=\left(\mathrm{b}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{5p}^{\mathrm{3}} −\mathrm{15pq}=\mathrm{13a}^{\mathrm{2}} \left(\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}\right)−\mathrm{2a}\left(\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\right)}\\{\mathrm{p}^{\mathrm{4}} −\mathrm{4p}^{\mathrm{2}} \mathrm{q}=\left(\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{80p}^{\mathrm{3}} −\mathrm{240pq}=\mathrm{8p}\left(\mathrm{13a}^{\mathrm{4}} −\mathrm{2a}^{\mathrm{3}} −\mathrm{26a}^{\mathrm{2}} \right)}\\{\mathrm{60p}^{\mathrm{4}} −\mathrm{240p}^{\mathrm{2}} \mathrm{q}=\mathrm{15}\left(\mathrm{a}^{\mathrm{4}} +\mathrm{4a}^{\mathrm{2}} +\mathrm{4}\right)}\end{cases} \\ $$$$\mathrm{140p}^{\mathrm{4}} −\mathrm{8pa}^{\mathrm{2}} \left(\mathrm{13a}^{\mathrm{2}} −\mathrm{2a}−\mathrm{26}\right)−\mathrm{15}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$

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