solve { ((x(√x)+y(√y) = 5)),((x(√y) +y(√x) = 1)) :}


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Question Number 111909 by john santu last updated on 05/Sep/20

$s o l v e {\begin{cases} x \sqrt{x} + y \sqrt{y} = 5 \\ x \sqrt{y} + y \sqrt{x} = 1 \end{cases}$
	Answered by bemath last updated on 05/Sep/20
	$let (√x) = u ; (√y) = v { ((u^3 +v^3 = 5⇒(u+v)^3 −3uv(u+v)=5)),((u^2 v+uv^2 = 1⇒uv(u+v)=1)) :} ⇔ (u+v)^3 −3.1 = 5 ⇒u+v = (8)^(1/(3 )) ⇒u+v = 2 ∧ u.v = (1/2) ⇒(√x) +(√(y )) = 2 ; (√(xy)) = (1/2)→xy =(1/4) ⇒((√x)+(√y))^2 = 4 ⇒ x+y+2(√(xy)) = 4 ⇒ x+y = 3→y=3−x now x(3−x) = (1/4) ; −x^2 +3x = (1/4) x^2 −3x+(1/4) = 0 ⇒x = ((3 ± (√(9−1)))/2) ⇒x = ((3±2(√2))/2) →y=3−(((3±2(√2) )/2)) y= ((3 ∓ 2(√2))/2) .$
	$l e t \sqrt{x} = u; \sqrt{y} = v$ ${\begin{cases} u^{3} + v^{3} = 5 \Rightarrow {(u + v)}^{3} - 3 u v (u + v) = 5 \\ u^{2} v + {u v}^{2} = 1 \Rightarrow u v (u + v) = 1 \end{cases}$ $\Leftrightarrow {(u + v)}^{3} - 3.1 = 5 \Rightarrow u + v = \sqrt[3]{8}$ $\Rightarrow u + v = 2 \land u . v = \frac{1}{2}$ $\Rightarrow \sqrt{x} + \sqrt{y} = 2; \sqrt{x y} = \frac{1}{2} \to x y = \frac{1}{4}$ $\Rightarrow {(\sqrt{x} + \sqrt{y})}^{2} = 4$ $\Rightarrow x + y + 2 \sqrt{x y} = 4$ $\Rightarrow x + y = 3 \to y = 3 - x$ $n o w x (3 - x) = \frac{1}{4}; - x^{2} + 3 x = \frac{1}{4}$ $x^{2} - 3 x + \frac{1}{4} = 0 \Rightarrow x = \frac{3 \pm \sqrt{9 - 1}}{2}$ $\Rightarrow x = \frac{3 \pm 2 \sqrt{2}}{2} \to y = 3 - (\frac{3 \pm 2 \sqrt{2}}{2})$ $y = \frac{3 \mp 2 \sqrt{2}}{2} .$
	Answered by ajfour last updated on 05/Sep/20
	$p^3 +q^3 =5 ( let (√x)=p , (√y)=q ) p^2 q+pq^2 =1 ⇒(p+q){(p+q)^2 −3pq}=5 & pq(p+q)=1 let pq=m , p+q=s s(s^2 −(3/s))=5 ⇒ s^3 =8 , s=p+q= 2, 2ω, 2ω^2 m=(1/s) = (1/2), (ω^2 /2) , (ω/2) x+y=p^2 +q^2 =s^2 −2m xy=p^2 q^2 =m^2 x, y = (s^2 /2)−m±(√(((s^2 /2)−m)^2 −m^2 )) (s^2 /2)−m=2−(1/2)=(3/2) , = 2ω^2 −(ω^2 /2)=((3ω^2 )/2) , = ((3ω)/2) . x, y = (3/2)±(√((9/4)−(1/4))) =(3/2)±(√2) , ((3/2)±(√2))ω , ((3/2)±(√2))ω^2$
	$p^{3} + q^{3} = 5 (l e t \sqrt{x} = p, \sqrt{y} = q)$ $p^{2} q + {p q}^{2} = 1$ $\Rightarrow (p + q) {{(p + q)}^{2} - 3 p q} = 5$ $& p q (p + q) = 1$ $l e t p q = m, p + q = s$ $s (s^{2} - \frac{3}{s}) = 5$ $\Rightarrow s^{3} = 8, s = p + q = 2, 2 ω, 2 ω^{2}$ $m = \frac{1}{s} = \frac{1}{2}, \frac{ω^{2}}{2}, \frac{ω}{2}$ $x + y = p^{2} + q^{2} = s^{2} - 2 m$ $x y = p^{2} q^{2} = m^{2}$ $x, y = \frac{s^{2}}{2} - m \pm \sqrt{{(\frac{s^{2}}{2} - m)}^{2} - m^{2}}$ $\frac{s^{2}}{2} - m = 2 - \frac{1}{2} = \frac{3}{2},$ $= 2 ω^{2} - \frac{ω^{2}}{2} = \frac{3 ω^{2}}{2},$ $= \frac{3 ω}{2} .$ $x, y = \frac{3}{2} \pm \sqrt{\frac{9}{4} - \frac{1}{4}}$ $= \frac{3}{2} \pm \sqrt{2}, (\frac{3}{2} \pm \sqrt{2}) ω, (\frac{3}{2} \pm \sqrt{2}) ω^{2}$
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