{ ((x+y(√x) = ((95)/8))),((y+x(√y) = ((93)/8))) :} . Find (√(xy))


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Question Number 107313 by bemath last updated on 10/Aug/20

${\begin{cases} x + y \sqrt{x} = \frac{95}{8} \\ y + x \sqrt{y} = \frac{93}{8} \end{cases} . F i n d \sqrt{x y}$
	Answered by Her_Majesty last updated on 10/Aug/20

	$a n s w e r e d$ $q u 106259$
	Answered by 1549442205PVT last updated on 26/Aug/20
	$Set (√x)+(√y)=a,(√(xy)) =b we have ((√x)−(√y))^2 =((√x)+(√y))^2 −4(√(xy))=a^2 −4b ⇒(√x)+(√y)=(√(a^2 −4b)). From that we get { (((√x)=((a+(√(a^2 −4b)))/2))),(((√y)=((a−(√(a^2 −4b)))/2))) :} ⇒ { ((x=((a^2 −2b+a(√(a^2 −4b)))/2))),((y=((a^2 −2b−a(√(a^2 −4b)))/2))) :} (∗) Substracting two equations we get x−y+(√(xy))((√y)−(√x))=(1/4) ⇔a(√(a^2 −4b)) −b(√(a^2 −4b)) =(1/4) ⇔(a−b)(√(a^2 −4b)) =(1/4) ⇔(a^2 −2ab+b^2 )(a^2 −4b)=(1/(16))(1) Adding up two equations we get x+y+(√(xy)) ((√x) +(√y) )=((47)/2) ⇔a^2 −2b+ab=((47)/2)⇒b=((2a^2 −47)/(4−2a))(2) ⇒b^2 =((4a^4 −188a^2 +2209)/(4a^2 −16a+16)).Replace into (1)we get [(a^2 −((4a^3 −94a)/(4−2a))+((4a^4 −188a^2 +2209)/(4a^2 −16a+16)).)(a^2 −((8a^2 −188)/(4−2a)))=(1/(16)) ⇒((4a^4 −16a^3 +16a^2 +8a^4 −16a^3 −188a^2 +376a+4a^4 −188a^2 +2209)/(4a^4 −16a+16))×((2a^3 +4a^2 −188)/(2a−4))=(1/(16)) After simple transformations we get 64a^7 −1696a^5 −7392a^4 +23875a^3 +153038a^2 −141376a−830576=0 ⇔(a−4)(64a^6 +256a^5 −672a^4 −10080a^3 −16445a^2 +87258a+207644)=0 ⇒a−4=0⇔a=4 We shall prove that the equation 64a^6 +256a^5 −672a^4 −10080a^3 −16445a^2 +87258a+207644=0 has no positive roots^((•)) (leave later) Replace into(2)we obtain b=((15)/4) substituting into (∗) we get x=((a^2 −2b+a(√(4a^2 −b)))/2)=((25)/4), y=((a^2 −2b−a(√(4a^2 −b)))/2)=(9/9) Thus,(√(xy))=b=15/4 Now we need have to prove that the eqn. 64x^6 +256x^5 −672x^4 −10080x^3 −17445x^2 +87258x+207644=0 has no positive roots Put x=(1/u) then our problem reduce to Prove that the equation 207644t^6 +87258t^5 −16445t^4 −10080t^3 −672t^2 +256t+64=0 has no positive roots .Indeed, Apply Cauchy′s inequality for positive numbers we have 64000t^6 +64000t^5 +64t≥3^3 (√(64^3 ×10^6 t^(12) ))=19200t^4 Hence,LHS≥143644t^6 +23258t^5 +2755t^4 −10080t^3 −672t^2 +192t+64 We have also 23258t^5 +2755t^4 ≥(√(23258.2755t))$
	$Set \sqrt{x} + \sqrt{y} = a, \sqrt{xy} = b we have$ ${(\sqrt{x} - \sqrt{y})}^{2} = {(\sqrt{x} + \sqrt{y})}^{2} - 4 \sqrt{xy} = a^{2} - 4 b$ $\Rightarrow \sqrt{x} + \sqrt{y} = \sqrt{a^{2} - 4 b} . From that we get$ ${\begin{cases} \sqrt{x} = \frac{a + \sqrt{a^{2} - 4 b}}{2} \\ \sqrt{y} = \frac{a - \sqrt{a^{2} - 4 b}}{2} \end{cases} \Rightarrow {\begin{cases} x = \frac{a^{2} - 2 b + a \sqrt{a^{2} - 4 b}}{2} \\ y = \frac{a^{2} - 2 b - a \sqrt{a^{2} - 4 b}}{2} \end{cases} ()$ $Substracting two equations we get$ $x - y + \sqrt{xy} (\sqrt{y} - \sqrt{x}) = \frac{1}{4}$ $\Leftrightarrow a \sqrt{a^{2} - 4 b} - b \sqrt{a^{2} - 4 b} = \frac{1}{4}$ $\Leftrightarrow (a - b) \sqrt{a^{2} - 4 b} = \frac{1}{4}$ $\Leftrightarrow (a^{2} - 2 ab + b^{2}) (a^{2} - 4 b) = \frac{1}{16} (1)$ $Adding up two equations we get$ $x + y + \sqrt{xy} (\sqrt{x} + \sqrt{y}) = \frac{47}{2}$ $\Leftrightarrow a^{2} - 2 b + ab = \frac{47}{2} \Rightarrow b = \frac{{2 a}^{2} - 47}{4 - 2 a} (2)$ $\Rightarrow b^{2} = \frac{{4 a}^{4} - {188 a}^{2} + 2209}{{4 a}^{2} - 16 a + 16} . Replace into$ $(1) we get$ $[(a^{2} - \frac{{4 a}^{3} - 94 a}{4 - 2 a} + \frac{{4 a}^{4} - {188 a}^{2} + 2209}{{4 a}^{2} - 16 a + 16} .) (a^{2} - \frac{{8 a}^{2} - 188}{4 - 2 a}) = \frac{1}{16}$ $\Rightarrow \frac{{4 a}^{4} - {16 a}^{3} + {16 a}^{2} + {8 a}^{4} - {16 a}^{3} - {188 a}^{2} + 376 a + {4 a}^{4} - {188 a}^{2} + 2209}{{4 a}^{4} - 16 a + 16} \times \frac{{2 a}^{3} + {4 a}^{2} - 188}{2 a - 4} = \frac{1}{16}$ $After simple transformations we get$ ${64 a}^{7} - {1696 a}^{5} - {7392 a}^{4} + {23875 a}^{3} + {153038 a}^{2} - 141376 a - 830576 = 0$ $\Leftrightarrow (a - 4) ({64 a}^{6} + {256 a}^{5} - {672 a}^{4} - {10080 a}^{3} - {16445 a}^{2} + 87258 a + 207644) = 0$ $\Rightarrow a - 4 = 0 \Leftrightarrow a = 4$ $We shall prove that the equation$ ${64 a}^{6} + {256 a}^{5} - {672 a}^{4} - {10080 a}^{3} - {16445 a}^{2} + 87258 a + 207644 = 0$ $has no positive roots^{(∙)} (leave later)$ $Replace into (2) we obtain b = \frac{15}{4}$ $substituting into () we get$ $x = \frac{a^{2} - 2 b + a \sqrt{{4 a}^{2} - b}}{2} = \frac{25}{4},$ $y = \frac{a^{2} - 2 b - a \sqrt{{4 a}^{2} - b}}{2} = \frac{9}{9}$ $Thus, \sqrt{xy} = b = 15 / 4$ $Now we need have to prove that the eqn .$ ${64 x}^{6} + {256 x}^{5} - {672 x}^{4} - {10080 x}^{3} - {17445 x}^{2} + 87258 x + 207644 = 0$ $has no positive roots$ $Put x = \frac{1}{u} then our problem reduce to$ $Prove that the equation$ ${207644 t}^{6} + {87258 t}^{5} - {16445 t}^{4} - {10080 t}^{3}$ $- {672 t}^{2} + 256 t + 64 = 0 has no positive roots$ $. Indeed,$ $Apply {Cauchy}^{'} s inequality for$ $positive numbers we have$ ${64000 t}^{6} + {64000 t}^{5} + 64 t ⩾ 3^{3} \sqrt{64^{3} \times 10^{6} t^{12}} = {19200 t}^{4}$ $Hence, LHS ⩾ {143644 t}^{6} + {23258 t}^{5} + {2755 t}^{4}$ $- {10080 t}^{3} - {672 t}^{2} + 192 t + 64$ $We have also$ ${23258 t}^{5} + {2755 t}^{4} ⩾ \sqrt{23258 . 2755 t}$
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