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Question Number 163627 by nurtani last updated on 08/Jan/22

	Answered by mr W last updated on 08/Jan/22
	$u=(1/(x+y)) >0 v=xy (√(21v))(1−u^2 )=8(√2) v=((128)/(21(1−u^2 )^2 )) x=(4/(3(1+u)^2 )) y=((32)/(7(1−u)^2 )) u=(1/(x+y))=(1/((4/(3(1+u)^2 ))+((32)/(7(1−u)^2 ))))=((21(1−u^2 )^2 )/(4[7(1−u)^2 +24(1+u)^2 ])) 4u=((21−42u^2 +21u^4 )/(31u^2 +34u+31)) 21u^4 −124u^3 −178u^2 −124u+21=0 (3u^2 −22u+3)(7u^2 +10u+7)=0 3u^2 −22u+3=0 u=((11−4(√7))/3) x+y=(1/u)=((11+4(√7))/3) xy=v=((128)/(21(1−u^2 )^2 ))=((466+176(√7))/(147)) x,y are roots of z^2 −((11+4(√7))/3)z+((466+176(√7))/(147))=0 x,y=(1/2){((11+4(√7))/3)±(√((((11+4(√7))/3))^2 −4(((466+176(√7))/(147)))))} x=(1/6){11+4(√7)−((5(√(233+88(√7))))/7)}≈1.0278 y=(1/6){11+4(√7)+((5(√(233+88(√7))))/7)}≈6.1666$
	$u = \frac{1}{x + y} > 0$ $v = x y$ $\sqrt{21 v} (1 - u^{2}) = 8 \sqrt{2}$ $v = \frac{128}{21 {(1 - u^{2})}^{2}}$ $x = \frac{4}{3 {(1 + u)}^{2}}$ $y = \frac{32}{7 {(1 - u)}^{2}}$ $u = \frac{1}{x + y} = \frac{1}{\frac{4}{3 {(1 + u)}^{2}} + \frac{32}{7 {(1 - u)}^{2}}} = \frac{21 {(1 - u^{2})}^{2}}{4 [7 {(1 - u)}^{2} + 24 {(1 + u)}^{2}]}$ $4 u = \frac{21 - 42 u^{2} + 21 u^{4}}{31 u^{2} + 34 u + 31}$ $21 u^{4} - 124 u^{3} - 178 u^{2} - 124 u + 21 = 0$ $(3 u^{2} - 22 u + 3) (7 u^{2} + 10 u + 7) = 0$ $3 u^{2} - 22 u + 3 = 0$ $u = \frac{11 - 4 \sqrt{7}}{3}$ $x + y = \frac{1}{u} = \frac{11 + 4 \sqrt{7}}{3}$ $x y = v = \frac{128}{21 {(1 - u^{2})}^{2}} = \frac{466 + 176 \sqrt{7}}{147}$ $x, y a r e r o o t s o f$ $z^{2} - \frac{11 + 4 \sqrt{7}}{3} z + \frac{466 + 176 \sqrt{7}}{147} = 0$ $x, y = \frac{1}{2} {\frac{11 + 4 \sqrt{7}}{3} \pm \sqrt{{(\frac{11 + 4 \sqrt{7}}{3})}^{2} - 4 (\frac{466 + 176 \sqrt{7}}{147})}}$ $x = \frac{1}{6} {11 + 4 \sqrt{7} - \frac{5 \sqrt{233 + 88 \sqrt{7}}}{7}} \approx 1.0278$ $y = \frac{1}{6} {11 + 4 \sqrt{7} + \frac{5 \sqrt{233 + 88 \sqrt{7}}}{7}} \approx 6.1666$
	Commented by nurtani last updated on 09/Jan/22

	$T h a n k y o u s i r$
	Commented by Tawa11 last updated on 09/Jan/22

	$Great sir$
	Answered by behi834171 last updated on 09/Jan/22

	$x \neq 0, y \neq 0$ $\frac{2}{\sqrt{3 x}} + \frac{4 \sqrt{2}}{\sqrt{7 y}} = 2 \Rightarrow \frac{2 \sqrt{2}}{\sqrt{7 y}} = 1 - \frac{1}{\sqrt{3 x}} \Rightarrow$ $\Rightarrow \frac{8}{7 y} = 1 - \frac{2}{\sqrt{3 x}} + \frac{1}{3 x} \Rightarrow y = \frac{8}{7 (1 - \frac{2}{\sqrt{3 x}} + \frac{1}{3 x})}$ $\Rightarrow \sqrt{3 x} (1 + \frac{1}{x + \frac{8}{7 (1 - \frac{2}{\sqrt{3 x}} + \frac{1}{3 x})}}) = 2 \overset{\frac{1}{\sqrt{3 x}} = m}{\Rightarrow}$ $\frac{1}{m} (1 + \frac{1}{\frac{1}{3 m^{2}} + \frac{8}{7 - 14 m + 7 m^{2}}}) = 2 \Rightarrow$ $\frac{1}{3 m^{2}} + \frac{8}{7 - 14 m + 7 m^{2}} = \frac{1}{2 m - 1} \Rightarrow$ $\frac{8}{7 - 14 m + 7 m^{2}} = \frac{3 m^{2} - 2 m + 1}{3 m^{2} (2 m - 1)} \Rightarrow$ $\Rightarrow 24 m^{2} (2 m - 1) = 7 (m^{2} - 2 m + 1) (3 m^{2} - 2 m + 1) \Rightarrow$ $\Rightarrow 48 m^{3} - 24 m^{2} = 7 (3 m^{4} - 8 m^{3} + 8 m^{2} - 4 m + 1) \Rightarrow$ $\Rightarrow 21 m^{4} - 104 m^{3} + 80 m^{2} - 28 m + 7 = 0$ $\Rightarrow m = 0.57, m = 4.1$ $\Rightarrow \frac{1}{3 x} = m^{2} \Rightarrow x = \frac{1}{3 m^{2}} \Rightarrow [x = 1.03, 0.021]$ $\Rightarrow \frac{8}{7 y} = 1 - 2 m + m^{2} \Rightarrow [y = 6.2, 0.12] . ◼$
	Commented by nurtani last updated on 09/Jan/22

	$t h a n k y o u s i r$
	Commented by mr W last updated on 09/Jan/22

	$o n l y o n e s o l u t i o n f u l f i l l s t h e$ $o r i g i n a l e q u a t i o n s!$
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